c
(c) Rate of cooling ${R_C} = \frac{{A\varepsilon \sigma ({T^4} - T_0^4)}}{{mc}}$$ = \frac{{A\varepsilon \sigma ({T^4} - T_0^4)}}{{V\rho C}}$
==> ${R_C} \propto \frac{A}{V} \propto \frac{1}{r} \propto \,\frac{1}{{({\rm{Diameter)}}}}$ $(\because \,m = \rho V)$
Since diameter of $A$ is half that of $B $ so it's rate of cooling will be doubled that of B