Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by $$\rho(r) = kr$$, 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:

First of all we determine the constant that appears in the charge-density expression. The charge density inside the solid sphere is given as $$\rho(r)=kr$$. The total charge present in the sphere is stated to be $$2Q$$, so we write

$$2Q=\int_{0}^{R}\rho(r)\,d\tau =\int_{0}^{R}kr\;4\pi r^{2}\,dr =4\pi k\int_{0}^{R}r^{3}\,dr =4\pi k\left[\frac{r^{4}}{4}\right]_{0}^{R} =4\pi k\left(\frac{R^{4}}{4}\right) =\pi kR^{4}.$$

Solving for $$k$$ gives

$$k=\frac{2Q}{\pi R^{4}}.$$

Now we focus on one of the two point charges, say charge A, which is situated at a distance $$a$$ from the centre. Because the volume charge distribution is spherically symmetric, Gauss’s law tells us that only the charge inside a Gaussian sphere of radius $$a$$ contributes to the electric field at A. The enclosed charge is

$$Q_{\text{enc}}(a)=\int_{0}^{a}\rho(r)\,d\tau =\int_{0}^{a}kr\;4\pi r^{2}\,dr =4\pi k\int_{0}^{a}r^{3}\,dr =4\pi k\left[\frac{r^{4}}{4}\right]_{0}^{a} =\pi k a^{4}.$$

Substituting the value of $$k$$ just obtained,

$$Q_{\text{enc}}(a)=\pi\left(\frac{2Q}{\pi R^{4}}\right)a^{4} =\frac{2Qa^{4}}{R^{4}}.$$

Next we write the electric field at A due to the spherically distributed charge. Gauss’s law in the form $$E(4\pi a^{2})=\dfrac{Q_{\text{enc}}}{\varepsilon_{0}}$$, or equivalently $$E=k_{e}\dfrac{Q_{\text{enc}}}{a^{2}}$$ with $$k_{e}=\dfrac{1}{4\pi\varepsilon_{0}}$$, gives

$$E_{\text{sphere}}=k_{e}\frac{Q_{\text{enc}}(a)}{a^{2}} =k_{e}\frac{2Qa^{4}/R^{4}}{a^{2}} =k_{e}\frac{2Qa^{2}}{R^{4}}.$$

The force exerted on charge A (whose value is $$-Q$$) by the sphere therefore has magnitude

$$F_{\text{sphere}} =Q\;E_{\text{sphere}} =Q\left(k_{e}\frac{2Qa^{2}}{R^{4}}\right) =2k_{e}\frac{Q^{2}a^{2}}{R^{4}}.$$

This force is directed towards the centre, because the sphere is positively charged while A is negative.

Now we evaluate the force on A due to the other point charge B. Charges A and B are both $$-Q$$ and are placed diametrically opposite, so their separation is $$2a$$. The Coulomb force magnitude is

$$F_{\text{B on A}} =k_{e}\frac{Q^{2}}{(2a)^{2}} =k_{e}\frac{Q^{2}}{4a^{2}}.$$

This force is directed away from the centre, i.e., outward along the line joining the two charges, because like charges repel.

According to the problem statement, charge A experiences no net force. Hence the outward force due to charge B must exactly cancel the inward force due to the sphere:

$$F_{\text{B on A}} = F_{\text{sphere}}$$ $$\Rightarrow\; k_{e}\frac{Q^{2}}{4a^{2}} =2k_{e}\frac{Q^{2}a^{2}}{R^{4}}.$$

We can divide both sides by the common factor $$k_{e}Q^{2}$$ to obtain

$$\frac{1}{4a^{2}} = \frac{2a^{2}}{R^{4}}.$$

Cross-multiplying gives

$$R^{4} = 8a^{4}.$$

Solving for $$a$$, we have

$$a^{4} = \frac{R^{4}}{8} \;\;\Longrightarrow\;\; a = \frac{R}{8^{1/4}} = 8^{-1/4}R.$$

This value exactly matches Option D.

Hence, the correct answer is Option D.