If we connect a high resistance($R$) in series with the galvanometer, it can be successfully converted into a voltmeter. Therefore,

Voltage drop across voltmeter:

$$\begin{gathered} V=I_g\left(R_g+R\right) \\ \Rightarrow 100=I_g\left(50+R\right) \end{gathered}$$

$\Rightarrow R=\frac{100}{5\times {10}^{-3}}-50$

$=19950 \Omega$