Maximum Charge Density Value for Zero Core Radius

To find the maximum charge density $d$ when $a = 0$, we analyze the given charge density distribution $\rho(r)$.

From the figure, the variation of charge density $\rho(r)$ with radial distance $r$ for $a \le r \le R$ is given by a straight line joining $(a, d)$ and $(R, 0)$:
$\rho(r) = d \left(1 - \frac{r - a}{R - a}\right)$

For $a = 0$, the charge density simplifies to:
$\rho(r) = d \left(1 - \frac{r}{R}\right) \quad \text{for } 0 \le r \le R$

The total charge $Q = Ze$ distributed within the nucleus of radius $R$ is obtained by integrating the charge over thin spherical shells of radius $r$ and thickness $dr$:
$Ze = \int_{0}^{R} \rho(r) \cdot 4\pi r^2 \, dr$

Substituting the expression for $\rho(r)$:
$Ze = 4\pi d \int_{0}^{R} \left(1 - \frac{r}{R}\right) r^2 \, dr$
$Ze = 4\pi d \int_{0}^{R} \left(r^2 - \frac{r^3}{R}\right) \, dr$

Evaluating the integral:
$Ze = 4\pi d \left[ \frac{r^3}{3} - \frac{r^4}{4R} \right]_{0}^{R}$
$Ze = 4\pi d \left( \frac{R^3}{3} - \frac{R^3}{4} \right)$
$Ze = 4\pi d \left( \frac{R^3}{12} \right)$
$Ze = \frac{\pi d R^3}{3}$

Solving for $d$:
$d = \frac{3Ze}{\pi R^3}$

Thus, the correct option is B.