Here,

$R= ?$

$G=\mathrm{50 }\mathrm{\Omega }$

$V=\mathrm{20 }\mathrm{V}$

$I_g=5\times 10^{-\mathrm{3 }}\mathrm{A}$

Where G: Resistance of the galvanometer

$V:$ Required range of the voltmeter

$I_g:$ Current in galvanometer which given full deflection

$R:$ High resistance in series with galvanometer

Applying the formula,

$R=\frac{V}{I_g}-G$

$=\left(\frac{20}{5\times {10}^{-4}}\right)-50$

$=4000-50$

$=3950 \mathrm{\Omega }$