Potential at the surface of spherical conductor of radius $\mathrm{R}$ carrying charge $\mathrm{Q},$ $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}}$

Let $Q_{1}$ and $Q_{2}$ are the charges on two spherical conductors of radii $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ respectively.

When these two charged spherical conductors connected by a wire. the potential at their surfaces becomes equal.

${\therefore \mathrm{V}_{\mathrm{s}_{1}}=\mathrm{V}_{\mathrm{s}_{2}} \Rightarrow \frac{\sigma_{1} \times 4 \pi \mathrm{R}_{1}^{2}}{4 \pi \varepsilon_{0} \mathrm{R}_{1}}=\frac{\sigma_{2} \times 4 \pi \mathrm{R}_{2}^{2}}{4 \pi \varepsilon_{0} \mathrm{R}_{2}}} $

${\Rightarrow \frac{\sigma_{1}}{\sigma_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}}$