Here, the current sensitivity is $10 \raisebox{1ex}{$\mathrm{div}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mA}$}\right.$ and there are $100$ divisions on the scale. So, the current required for full-scale deflection is

$I_g=\frac{1}{10}\times 100 \mathrm{mA}=10 \mathrm{mA}$

$= 0.01 \mathrm{A}$

Also, the voltage sensitivity is $2\mathrm{ }\raisebox{1ex}{$\mathrm{div}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mV}$}\right.$ so the voltage required for full-scale deflection is

$V_g=\frac{1}{2}\times 100\mathrm{ }\mathrm{mV}$

$= 0.05 \mathrm{V}$

Galvanometer resistance is given by

$G=\frac{V_g}{I_g}=\frac{0.05}{0.01}$

$=5 \mathrm{\Omega }$