Let charges on metal spheres before contact are $\mathrm{Q}_{1}$ and $\mathrm{Q}_{2}$ hence,

$\mathrm{Q}_{1}=\sigma \times 4 \pi \mathrm{R}^{2} \text { and } \mathrm{Q}_{2}=\sigma \times 4 \pi(2 \mathrm{R})^{2} ~\\ =\sigma \times 16 \pi \mathrm{R}^{2}$ $=4 \mathrm{Q}_{1}$

Let the charges on metal sphere, after coming in contact becomes $\mathrm{Q}_{1}{ }^{\prime}$ and $\mathrm{Q}_{2}{ }^{\prime} .$

According to law of conservation of charges,

$\mathrm{Q}_{1}^{\prime}+\mathrm{Q}_{2}^{\prime}=\mathrm{Q}_{1}+\mathrm{Q}_{2}$ $=\mathrm{Q}_{1}+4 \mathrm{Q}_{1}$ $=5 \mathrm{Q}_{1}$ $=5\left(\sigma \times 4 \pi \mathrm{R}^{2}\right)$

When metal spheres come in contact they acquire equal potentials.

$\therefore \mathrm{V}_{1}=\mathrm{V}_{2}$

$\frac{k \mathrm{Q}_{1}^{\prime}}{\mathrm{R}}=\frac{k \mathrm{Q}_{2}^{\prime}}{\mathrm{R}}$

Putting value of $Q_{1}^{\prime}$ and $Q_{2}^{\prime}$ in above equations,

$\mathrm{Q}_{1}=\frac{5}{3}\left(\sigma \times 4 \pi \mathrm{R}^{2}\right)$

$text { and } \mathrm{Q}_{2}=\frac{10}{3}\left(\sigma \times 4 \pi \mathrm{R}^{2}\right)$

$\therefore \sigma_{1}=\frac{5}{3} \sigma \text { and } \sigma_{2}=\frac{5}{6} \sigma$