You did catch a good segue to the problem, with the \$3.3\:\textrm{V}-700\:\textrm{mV}=2.6\:\textrm{V}\$. But to continue, you have to realize that this is the voltage siting across the resistor at the base. With \$3.3\:\textrm{V}\$ on the left side and \$700\:\textrm{mV}\$ on the right, it has to be that way. If you know Ohm's law, then you know that \$I=\frac{V}{R}\$ and from this you can compute the current that must be going into the transistor base (where else would the current go?) However, you are supposed to figure out that resistor value. So stop there for a moment. Time to look elsewhere and see if there is something useful to add.
You are told that the LED requires \$100\:\textrm{mA}\$ of current. That would seem to be another bit of data and probably an important one. This current must go through both \$R_C\$ and also the transistor collector. At this point it is a good idea to consider some reality checks.
Go take a look at this datasheet for the 2N2222A. (That is actually what they used to look like, by the way, with a small metal TO-18 can. I still have a bunch of them. But they are comparatively more expensive that the TO-92 packaged PN2222A and so are not used so much anymore.) You will indeed see that the total dissipation of the device is given as \$1\:\textrm{W}\$. However, that assumes that you can keep the case at room temperature. Good luck with that. It's better to look at the "Thermal Resistance, Junction to Ambient" and see that this is \$325\:\frac{^\circ\textrm{C}}{\textrm{W}}\$. This is what it would be without any extra heat sinks and assuming reasonable and open exposure to air that can move and circulate. Burning a full watt would suggest a rise in temperature of \$325\:^\circ\textrm{C}\$ and that probably isn't such a good thing. Note that they only rate it as \$\frac{1}{2}\:\textrm{W}\$ in this case, too. All this together says that the device can still work at about \$175\:^\circ\textrm{C}\$ at the junction. But that's also still not a good idea.
Back to the design issues. You have a \$5\:\textrm{V}\$ power source for the LED. (Luckily well under the "breakdown voltage" for it.) The transistor should be operated in "saturated" mode (which often just means "as a switch") Let's go to the following figure on that datasheet:
Here, you can see a very conveniently placed curve described as \$100\:\textrm{mA}\$! You will want a very low \$V_{CE_{SAT}}\$ voltage (to avoid wasting power in the transistor and so as to provide as much remaining voltage for the LED and resistor.) However, you don't need to go crazy here. Just look for the knee in the curve and go a little bit beyond it. (These are "typical" curves so you can't be certain any particular part will follow that curve perfectly.) You can see it diving downward and then sloping lower. It looks very much like \$10\:\textrm{mA}\$ might be a good choice for the base current, yes? You could go more, of course. But this looks pretty good to me.
So, now you have \$I_B=10\:\textrm{mA}\$ for \$I_C=100\:\textrm{mA}\$ and you can work out the value for your base resistor, finally. From the above chart, you can see that you need to plan on (we need to keep in mind "typical" here) about \$V_{CE_{SAT}}=200\:\textrm{mV}\$. The only remaining thing is to work out your collector resistor by working out the voltage across it. Here, I think you have a problem. You write that the LED has a voltage drop of \$700\:\textrm{mA}\$ at \$100\:\textrm{mA}\$. But that isn't reasonable. I think you need to look up your problem again. Regardless, you can subtract your LED voltage from \$5\:\textrm{V}\$ and then subtract \$V_{CE_{SAT}}\$ from that and the remaining voltage will be the drop you need across your collector resistor. So you can then calculate that value, finally, as well.
That should help you see how to proceed, once you get a reasonable figure for your LED voltage. And once you've completed the design, it should be easy to work out the power required for your resistors (either \$\frac{V_R^2}{R}\$ or else \$I_R^2 R\$.)
