Explanation

(b) Joined by a wire means they are at the same potential
For same potential $\frac{kQ_{1}}{a_{1}} = \frac{kQ_{2}}{a_{2}} \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{a}{b}$
Further, the electric field at the surface of the sphere having radius $R$ and charge $Q$ is $\frac{kQ}{R^{2}}$
$\therefore\frac{E_{1}}{E_{2}} = \frac{kQ_{1}/a^{2}}{kQ_{2}/b^{2}} = \frac{Q_{1}}{Q_{2}} \times \frac{b^{2}}{a^{2}} = \frac{b}{a}$