Copper's density at \$20\,^\circ\$C temperature is \$\rho_\text{Cu}\approx 8.96\,\frac{\text{g}}{\text{cc}}\$. It's atomic mass is \$m_\text{Cu}\approx 63.546\,\text{u}\$. Avogadro's constant is \$N_A=6.02214076\times 10^{23}\:\frac{1 }{mol}\$ (due to become the standard value in 2019.) You can compute from this the number of atoms in \$1\:\text{cc}\$ of copper as \$\frac{N_A\,\cdot\, \rho_\text{Cu}}{m_\text{Cu}}\approx 8.49\times 10^{22}\,\frac{\text{atoms}}{\text{cc}}\$. If each atom were able to donate one conduction band electron, then you'd expect that many conduction band electrons per cc, as well. So we'd estimate \$8.49\times 10^{22}\,\frac{\text{electrons}}{\text{cc}}\$.

There is a slight modification to this made by Fermi-Dirac statistics. Based upon the Fermi energy of copper, which is \$7.00\:\text{eV}\$, an integration is performed over the product of the electron state density term and the Fermi-Dirac distribution term. The result is \$\left[\frac{\pi\,8\sqrt{2\,m_e^3}}{h^3}\right]\cdot\bigg[\frac23\sqrt{E_{F_\text{Cu}}^3}\bigg]\approx 8.411\times 10^{22}\,\frac{\text{electrons}}{\text{cc}}\$.

Note that this is slightly lower than the more simplistic assumption. But not so far apart that the simplistic assumption isn't a practical one. (Don't forget the above is for pure copper, which isn't actually used in most copper wire you will find. But at least there is some consistency and we can say that 99.1% of the copper atoms donate a conduction band electron.)

With a \$10\:\text{V}\$ supply and a requirement of \$10\:\Omega\$ (a 20-gauge wire that is \$300.3\:\text{m}\$ long) to get \$1\:\text{A}\$, the electric field intensity must be only about \$33.3\:\frac{\text{mV}}{\text{m}}\$. Since conduction band electron mobility is the drift velocity divided by the electric field intensity, you can easily work out that the copper mobility figure is \$\mu_\text{Cu}\approx 4.3\times 10^{-3}\:\frac{\text{m}^2}{\text{V}\cdot\text{s}}\$.

Copper's density at \$20\,^\circ\$C temperature is \$\rho_{_\text{Cu}}\approx 8.96\,\frac{\text{g}}{\text{cc}}\$. It's atomic mass is \$m_{_\text{Cu}}\approx 63.546\,\text{u}\$. Avogadro's constant is \$N_A=6.02214076\times 10^{23}\:\text{mol}^{-1}\$ (standardized in 2019.)

The number of atoms in \$1\:\text{cc}\$ of copper is then \$\frac{N_A\,\cdot\, \rho_{_\text{Cu}}}{m_{_\text{Cu}}}\approx 8.49\times 10^{22}\,\frac{\text{atoms}}{\text{cc}}\$. Assuming each atom of copper is able to donate one conduction band electron, then you'd expect the same number of conduction band electrons per cc. So we'd estimate \$8.49\times 10^{22}\,\frac{\text{electrons}}{\text{cc}}\$ in the conduction band.

There is a slight modification to this made by Fermi-Dirac statistics. Based upon the Fermi energy of copper, which is \$E_{F_{_\text{Cu}}}=7.00\:\text{eV}\$, an integration is performed over the product of the electron state density term and the Fermi-Dirac distribution term. The result is:

$$\left[\frac{8\pi\,\sqrt{2\,m_e^3}}{h^3}\right]\cdot\bigg[\frac23\sqrt{E_{F_{_\text{Cu}}}^{\:\:3}}\bigg]\approx 8.411\times 10^{22}\,\frac{\text{electrons}}{\text{cc}}$$

Note that this is slightly lower than the more simplistic assumption. But not so far apart that the simplistic assumption isn't a practical one. (Don't forget the above is for pure copper, which isn't actually used in most copper wire you will find. But at least there is some consistency and we can say that 99.1% of the copper atoms donate a conduction band electron.)

Normally, in copper we might expect the orbital shells to fill up in the usual order of energy: 1s, then 2s, then 2p, then 3s, then 3p, then 4s, and then back to 3d. (The 4s is lower in energy than 3d.) Simplistically following that approach, we'd write: \$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^9\$. That gets all 29 electrons placed. But it's wrong. There's a special stability that occurs when shell levels are completely filled up and in this case the 3d (which holds 5 pairs, or a total of 10 electrons) prefers to fill up completely. (It sums out to a lower average energy using 3d this way, than leaving 3d slightly shy of being full.) So the result is that copper is actually: \$1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\$. It is this \$4s^1\$ electron that copper donates to the conduction band.

With a \$10\:\text{V}\$ supply and a requirement of \$10\:\Omega\$ (a 20-gauge wire that is \$300.3\:\text{m}\$ long) to get \$1\:\text{A}\$, the electric field intensity must be only about \$33.3\:\frac{\text{mV}}{\text{m}}\$. Since conduction band electron mobility is the drift velocity divided by the electric field intensity, you can easily work out that the copper mobility figure is \$\mu_{_\text{Cu}}\approx 4.3\times 10^{-3}\:\frac{\text{m}^2}{\text{V}\cdot\text{s}}\$.

The number of electrons involved on the surface is quite small. The calculations are a little more involved than I want to belabor here, but for the figure of \$33.3\:\frac{\text{mV}}{\text{m}}\$ indicated above in the 20-gauge wire example the number of electrons in the first centimeter's wire surface might be on the order of perhaps 1000 electrons or so. Completely negligible compared to the number of electrons in the copper found in that first centimeter (more than \$4\times 10^{20}\$.)

The number of excess electrons involved on the surface is quite small. The calculations are a little more involved than I want to belabor here, but for the figure of \$33.3\:\frac{\text{mV}}{\text{m}}\$ indicated above in the 20-gauge wire example the number of electrons in the first centimeter's wire surface might be on the order of perhaps 1000 electrons or so. Completely negligible compared to the number of electrons in the copper found in that first centimeter (more than \$4\times 10^{20}\$.)