Let a total charge $2Q$ be distributed in a sphere of radius $R$, with the charge density given by $\rho \left(r\right)=kr$, where $r$ is the distance from the centre. Two charges $A$ and $B$, of $-Q$ each, are placed on diametrically opposite points, at equal distance $a$, from the centre. If $A$ and $B$ do not experience any force, then

The correct option is C $a=8^{\left(\frac{-1}{4}\right)}R$
Consider an imaginary sphere of radius $a$ and concentric with the charged sphere.

According to Gauss' law,
$\oint \overrightarrow{E}.d\overrightarrow{A}=\frac{q_{encl}}{{\epsilon }_0}$



So, $E\times 4\pi r^2=\frac{1}{{ϵ}_0}\int \rho \left(4\pi r^2\right)dr$

$E\times 4\pi r^2=\frac{1}{{ϵ}_0}{\int }_0^r\left(kr\right)\left(4\pi r^2\right)dr$

$E\times 4\pi r^2=\frac{4\pi k}{{ϵ}_0}\left(\frac{r^4}{4}\right)$
$\therefore E=\frac{k}{4{ϵ}_0}{r}^2..........\left(i\right)$

Also, $2Q={\int }_0^R\left(kr\right)\left(4\pi r^2\right)dr$

$=4\pi k{\left|\frac{r^4}{4}\right|}_0^R=k\pi R^4$
$\therefore k=\frac{2Q}{\pi R^4}..........\left(ii\right)$

From equations $\left(i\right)$ and $\left(ii\right)$, we get,
$E=\frac{Q{r}^2}{2\pi {ϵ}_0{R}^4}..........\left(iii\right)$

According to the given condition, $F_{net}=0$
$\Rightarrow \frac{Q^2}{4\pi {ϵ}_0(2a{)}^2}=EQ$

Substituting $E$ from equation $\left(iii\right)$, we get,
$\Rightarrow \frac{Q}{16\pi {ϵ}_0{a}^2}=\frac{Q{a}^2}{2\pi {ϵ}_0{R}^4}$

After solving, we get,
$a=8^{\left(\frac{-1}{4}\right)}R$.

Hence, $\left(A\right)$ is the correct answer.