Let potential source (cell) of $V$ volts is connected across $B$ and $C$. Then potential on surface of $B=V_B=\frac{k q}{b}+\frac{k(-q)}{c}$

As given, $\quad V_B=V$

So, $\quad V=k q\left(\frac{1}{b}-\frac{1}{c}\right)$

$=k q\left(\frac{c-b}{b c}\right)$

$\Rightarrow \quad q=\frac{V b c}{k(c-b)}$

So, charge on $C$ is $-q=\frac{-V b c}{k(c-b)}$ where, $k=\frac{1}{4 \pi \varepsilon_0}=$ constant.

Hence, the charge on the sphere is

$-4 \pi \varepsilon_0 \frac{V b c}{(c-b)}$