Given, the resistance of the galvanometer $G=100\mathrm{\Omega }$
No. of divisions $=50$
Current Sensitivity  $=20\mathrm{\mu }\mathrm{A}/\mathrm{d}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{o}\mathrm{n}$

The maximum current through galvanometer will be 
$I_{max}=50\times 20\times {10}^{-6}={10}^{-3}\mathrm{A}$
For $V_1 = 2 V$

using Ohm's law $V_1= I_{max }\left(G+R_1\right)$
$2={10}^{-3} \times \left(100+R_1\right)$
$\therefore R_1=1900\mathrm{\Omega }$

For $V_2 = 10 V$

using Ohm's law $V_2 =I_{max}\left(G+R_1+R_2\right)$

$10={10}^{-3} \times \left(100 + 1900+R_2\right)$
$\therefore R_2=8000\mathrm{\Omega }$

For $V_3 =20 V$

using Ohm's law $V_3 =I_{max }\left(G+R_1+R_2+R_3\right)$
$20={10}^{-3} \left(100 +1900+8000+R_3\right)$
$\therefore R_3=10\times {10}^3\mathrm{\Omega }$