Let the resistance of the voltmeter be $R \mathrm{\Omega }$. The voltmeter is in parallel with the $5 \mathrm{\Omega }$. So the total resistance in the circuit is 

$\left(\frac{5R}{5+R}+2\right) \mathrm{\Omega }$

Applying Ohm's law in the whole circuit, $V=IR$,

$I=\left(\frac{3 }{2+{\displaystyle \frac{5R}{5+R}}}\right) \mathrm{A}$

It is given that the potential difference across $5 \mathrm{\Omega }$ is $2 \mathrm{V}$.

Thus, by Ohm's law,

$$\begin{gathered} \left(\frac{3}{2+{\displaystyle \frac{5R}{5+R}}}\right)\times \left(\frac{5R}{R+5}\right)=2 \\ \Rightarrow \frac{15+3R}{10+7R}\times \frac{5R}{5+R}=2 \\ \Rightarrow 15R=20+14R \\ \Rightarrow R=20 \mathrm{\Omega } \end{gathered}$$