Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges in them, repel each other with a force $F$ when kept apart at some distance. $A$ third spherical conductor $A$ having same radius as that of $B$ but uncharged, is brought in contact with $B$, then brought in contact with $C$ and finally removed away from both. The new force of repulsion between $B$ and $C$ is

Let the spherical conductors $B$ and $C$ have same charge $q$. The electric force between them is 

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$ where, $r$, 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 brought in contact with $C$, then charge on each conductor

$q_A=q_C=\frac{q_A+q_C}{2}=\frac{\left(q/2\right)+q}{2}=\frac{3q}{4}$

Hence, electric force acting between $B$ and $C$ is

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_B{q}_C}{r^2}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(q/2\right)\left(3q/4\right)}{r^2}=\frac{3}{8}\left[\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}\right]=\frac{3F}{8}$