In the figure, the inner (shaded) region A represents a sphere of...

Since both the densities are positive, the total charge can not be zero. Hence, option A is incorrect.

For total charge,

${Q}_{Total} = {∫}_{0}^{{r}_{A}}kr\left(4\pi{r}^{2}\right)dr + {∫}_{{r}_{A}}^{{r}_{B}}\dfrac{2k}{r}\left(4\pi{r}^{2}\right)dr$

$= \dfrac{4\pik}{4}{r}_{A}^{4} + \dfrac{8\pik}{2}\left({r}_{B}^{2} - {r}_{A}^{2}\right)$

$= \pik + 4\pik\left({r}_{B}^{2} - {r}_{A}^{2}\right)$

If ${r}_{B} = \dfrac{3}{2}$

${Q}_{Total} = \pik + 4\pik\left(\dfrac{9}{4} - 1\right)$

$= \pik + 4\pik\left(\dfrac{5}{4}\right) = 6\pik$

The potential just outside $B$ will be,

$V = \dfrac{1}{4\pi{ε}_{0}}\dfrac{{Q}_{total}}{{r}_{B}} = \dfrac{1}{4\pi{ε}_{0}}\dfrac{6\pik}{{r}_{B}} = \dfrac{3k}{2}\dfrac{2}{3{ε}_{0}} = \dfrac{k}{{ε}_{0}}$

Hence, option B is correct.

If ${r}_{B} = 2$

${Q}_{Total} = \pik + 4\pik\left(4 - 1\right)$= 13\pik$</p><p>Hence, option C is incorrect</p><p>If${r}_{B} = \dfrac{5}{2}$</p><p>${Q}_{Total} = \pik + 4\pik\left(\dfrac{25}{4} - 1\right)$</p><p>$= \pik + \pik\left(21\right)$</p><p>$= 22\pik$</p><p>Therefore, the electric field just outside B will be,</p><p>$E = \dfrac{1}{4\pi{ε}_{0}}\dfrac{{Q}_{Total}}{{{r}_{B}}^{2}}\dfrac{1}{4\pi{ε}_{0}}\dfrac{22\pik}{25} \times 4$$= \dfrac{22k}{25{ε}_{0}}

Hence, option D is incorrect.