A spherical metal shell A of radius R_A and a solid metal sphere B...

Inside a conducting shell, electric field is always zero. Therefore, option (a) is correct. When the two are connected, their potentials become the same. $$\begin{array}{ll} \therefore& V_A=V_B \\ \text{ or }& \frac{Q_A}{R_A}=\frac{Q_B}{R_B} \quad\left(\becauseV=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\right) \end{array}$$Since, $R_A \gt R_B$ \therefore \quad Q_A>Q_B $\therefore$ Option (b) is correct. Potential is also equal to, $$\begin{array}{rlrl}V & =\frac{\sigma R}{\varepsilon_0}\\ V_A & =V_B \\ \therefore \quad \sigma_A R_A & =\sigma_B R_B \\ \text{ or } \quad \sigma_B & \frac{\sigma_A}{\sigma_B}& =\frac{R_B}{R_A} \end{array}$$\therefore$Option (c) is correct. Electric field on surface, E=$\frac{\sigma}{E_0} \text{ or }$E \propto$\sigma$or$\quad \sigma_A \lt \sigma_B$Since,$\quad \sigma_A \lt \sigma_B\therefore \quad E_A \lt E_B\therefore$Option (d) is also correct.$\therefore$Correct options are (a), (b), (c) and (d). Analysis of Question (i) Question is simple. (ii)$E=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2}=\frac{\sigma}{\varepsilon_0}$(On surface) As,$\sigma=$surface charge density =$\frac{Q}{4 \pi R^2} ($iii)$V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}=\frac{\sigma R}{\varepsilon_0}$