First off, I removed the battery symbol. It's enough to specify a ground (you get to pick any one wire [aka: node] and call it \$0\:\textrm{V}\$) and the source voltage point.
Second, it doesn't matter which of the two \$35\:\textrm{k}\Omega\$ resistors you measure across (you should get the same answer either way), so I placed one end of the voltmeter at the "ground" (zero) reference point and the other end at an "interesting" point. If you look this over, you should be able to see that it asks the same question as does your newly added schematic.
Third, I've separated out the internal meter resistance as a discrete, added resistor to the schematic. The idea here is that now the voltmeter has "infinite" resistance and so it does not affect the circuit, anymore. But to keep the behavior the same, I had to add \$R_{METER}\$. Given your own schematic, I think you can see why this would still be the same question.

simulate this circuit – Schematic created using CircuitLab
Now, you should know how to calculate parallel resistances. So you can compute the equivalent single resistor you could use to replace the pair of \$R_2\$ and \$R_{METER}\$.
But before I go there, you can look at the resistors as instead being conductors. (In electronics, the symbol \$G\$ is used instead of \$R\$.) So \$G_1=\frac{1}{R_1}\$, \$G_2=\frac{1}{R_2}\$ and \$G_{METER}=\frac{1}{R_{METER}}\$. The total conductance of \$G_1\$ and \$G_{METER}\$ is just the sum of the two, because adding another conductance makes the whole thing more "conductive," right? So, if you add the conductance of \$R_2\$ and \$R_{METER}\$ and then convert that paired conductance back into a resistance again, you'd have the result of this as the paired resistance value of the two:
$$R_{METER}\mid\mid R_2=\frac{1}{\frac{1}{R_{METER}}+\frac{1}{R_2}}=\frac{R_{METER}\cdot R_2}{R_{METER}+R_2}=12.\overline{72}\:\textrm{k}\Omega$$
Now, it's just a voltage divider made up of two resistors. \$R_1\$ is still the same, but you now have a new replacement resistor that replaces the pairing of \$R_2\$ and \$R_{METER}\$. (Above computed value.)
From this divider, you should be able to calculate the resulting voltage.
Note that \$R_{TOTAL}=R_1+\left(R_2\mid\mid R_{METER}\right)\$ and that if you compute \$I_{TOTAL}\$ from that and \$V_{TOTAL}=10\:\textrm{V}\$, that this will NOT tell you the current through \$R_2\$. That's because \$R_2\$ must share this total current (all of which does have to flow through \$R_1\$) with \$R_{METER}\$.
An entirely different approach would be to construct the Thevenin equivalent before attaching the meter to it. This would be:

simulate this circuit
Once again, you have a new voltage divider circuit that will yield the exact same result. But a different approach to getting there.
Note here that a different \$R_{TOTAL}=\left(R_1\mid\mid R_2\right)+R_{METER}\$ is computed and that if you then compute a different \$I_{TOTAL}\$ from this (and \$V_{THEVENIN}=5\:\textrm{V}\$), that this will actually tell you the current through \$R_{METER}\$. That's because while \$R_1\$ and \$R_2\$ must share this total current, all of it does have to flow through \$R_{METER}\$. So you can compute the voltage by multiplying this current times \$R_{METER}\$.