Answer
(a) $V_1 = V_2$
(b) $Q_1 \gt Q_2$
(c) $E_1 \lt E_2$
Work Step by Step
(a) When a conductor is in equilibrium, all parts of the conductor has the same constant electric potential. Therefore, $V_1 = V_2$
(b) We can find an expression for the charge:
$V = \frac{k~Q}{R}$
$Q = \frac{V~R}{k}$
Since $R_1 \gt R_2$, then $Q_1 \gt Q_2$
(c) We can write an expression for the electric field at the surface:
$E = \frac{k~Q}{R^2}$
$E = \frac{k~V~R}{k~R^2}$
$E = \frac{V}{R}$
Since $R_1 \gt R_2$, then $E_1 \lt E_2$
