Explanation

(d) Let the spherical conductors $B\ and\ C\ have\ same\ charge\ as\ q.The\ $electric force between them is
$F = \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{r^{2}}$
$r,$ being the distance between them.
When third uncharged conductor $A$ is brought in contact with $B,$ then charge on each conductor
$q_{A} = q_{B} = \frac{q_{A} + q_{B}}{2}$
=$\frac{0 + q}{2} = \frac{q}{2}$
When this conductor $A$ is now brought in contact with $C$, then charge on each conductor
$q_{A} = q_{C} = \frac{q_{A} + q_{c}}{2}$
$= \frac{\left( \frac{q}{2} \right) + q}{2} = \frac{3q}{4}$
Hence, electric force acting between $B$ and $C$ is
$F^{'} = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{B}q_{c}}{r^{2}}\ $
$F^{'} = \frac{1}{4\pi\varepsilon_{0}\ }\frac{(q/2)(3q/4)}{r^{2}}\ $
$= \frac{3}{8}\left\lbrack \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{r^{2}} \right\rbrack = \frac{3F}{8}$