Explanation

(b) Potential at the surface of spherical conductor of radius $R$ carrying charge $Q,V_{S} = \frac{Q}{4\pi\varepsilon_{0}R}$
Let $Q_{1}$ and $Q_{2}$ are the charges on two spherical conductors of radii $R_{1}$ and $R_{2}$ respectively
When these two charged spherical conductors connected by a wire, the potential at their surfaces becomes equal
$\therefore Vs_{1} = Vs_{2}$
$\frac{Q_{1}}{4\pi\varepsilon_{0}R_{1}} = \frac{Q_{2}}{4\pi\varepsilon_{0}R_{2}} \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{R_{1}}{R_{2}}$ …(i)
Surface charge density, $\sigma = \frac{Charge}{Area}$
$\therefore\frac{\sigma_{1}}{\sigma_{2}} = \frac{\left( \frac{Q_{1}}{4\pi R_{1}^{2}} \right)}{\left( \frac{Q_{2}}{4\pi R_{2}^{2}} \right)} = \frac{Q_{1}}{Q_{2}} \times \frac{R_{2}^{2}}{R_{1}^{2}}$
$= \frac{R_{1}}{R_{2}} \times \frac{R_{2}^{2}}{R_{1}^{2}} = \frac{R_{2}}{R_{1}}$ [Using (i)]