1. Interpret the graph for $a=0$

   The figure described is the standard rho(r) vs $r$ graph where, for $a=0$, the charge density decreases linearly from its maximum value $d$ at $r=0$ to zero at $r=R$.

   Hence,

   $$$$

ho(r)=d\left(1-\frac{r}{R}\right), \qquad 0\le r\le R$$

2. Use total charge of nucleus

   The total nuclear charge is $Ze$, so $\int_V \rho(r)\, dV = Ze$

   In spherical coordinates, $dV=4\pi r^2dr$

   Therefore, $4\pi \int_0^R \rho(r) r^2 dr = Ze$

3. Substitute $\rho(r)$

   $4\pi \int_0^R d\left(1-\frac{r}{R}\right)r^2dr = Ze$

   $4\pi d\int_0^R \left(r^2-\frac{r^3}{R}\right)dr = Ze$

4. Evaluate the integral

   $\int_0^R r^2dr=\frac{R^3}{3}, \qquad \int_0^R r^3dr=\frac{R^4}{4}$

   So,

   = \frac{R^3}{3}-\frac{1}{R}\cdot \frac{R^4}{4} = R^3\left(\frac13-\frac14\right) = \frac{R^3}{12}$$ Hence, $$4\pi d\cdot \frac{R^3}{12}=Ze$$ $$\frac{\pi d R^3}{3}=Ze$$

5. Solve for $d$

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

6. Match with options

   This corresponds to: $\boxed{\text{B } \frac{3Ze}{\pi R^3}}$

7. Comparison with stored answer

   Stored correct answer: B

   Our derived answer also gives B, so they agree.