1. Initial charges and force

Let the initial charges on conductors $B$ and $C$ be equal: $q_B=q_C=q$

They are separated by a fixed distance $r$, so the initial force of repulsion is $F = k\frac{q\cdot q}{r^2}=k\frac{q^2}{r^2}$

where $k=\dfrac{1}{4\pi\varepsilon_0}$.

2. Third uncharged identical conductor touches $B$

Let the third conductor be $A$, initially uncharged. Since $A$ and $B$ are identical spheres, when they touch, total charge shared equally is: $q+0=q$ So after contact: $q_B=\frac{q}{2},\qquad q_A=\frac{q}{2}$

3. Now third conductor touches $C$

Before this contact:

• charge on $A$ is $\dfrac{q}{2}$
• charge on $C$ is $q$

Total charge on $A$ and $C$: $\frac{q}{2}+q=\frac{3q}{2}$

Again, since they are identical spheres, this charge is shared equally: $q_A=q_C=\frac{1}{2}\left(\frac{3q}{2}\right)=\frac{3q}{4}$

Thus after removal of $A$: $q_B=\frac{q}{2},\qquad q_C=\frac{3q}{4}$

4. New force between $B$ and $C$

The new repulsive force is $F' = k\frac{\left(\frac{q}{2}\right)\left(\frac{3q}{4}\right)}{r^2}$ $F' = k\frac{3q^2}{8r^2}$

Using $F=k\frac{q^2}{r^2}$ we get $F' = \frac{3}{8}F$

5. Correct option

$\boxed{\frac{3F}{8}}$ So the correct option is D.