(a) Consider a sphere $S$ of radius $R$ and two hypothetic sphere of radius $r < R$ and $r>R$.

$$
\text { According to Gauss's law, } \oint \text { E.dS }=\frac{\mathrm{q}}{\varepsilon_0}
$$
Case I : Consider, for Point $\mathrm{r} < \mathrm{R}$, electric field intensity will be
$$
\oint \mathrm{E} . \mathrm{dS}=\frac{1}{\varepsilon_0} \int \rho \mathrm{dV} \quad(\mathrm{q}=\rho \mathrm{dV})
$$
Volume of Gaussion surface $(\mathrm{V})$
As we know that,
For $\mathrm{dV}, \mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3 \Rightarrow \mathrm{dV}=3 \times \frac{4}{3} \pi \mathrm{r}^2 \mathrm{dr}=4 \pi \mathrm{r}^2 \mathrm{dr}$
$$
[\rho(r)=k r \text { for }(r < R)]
$$
So, $\quad \oint \mathrm{E} . \mathrm{dS}=\frac{1}{\varepsilon_0} 4 \pi \mathrm{K} \int_0^{\mathrm{r}} \mathrm{r}^3 \mathrm{dr}$
(E) $4 \pi \mathrm{r}^2=\frac{4 \pi \mathrm{K}}{\varepsilon_0} \frac{\mathrm{r}^4}{4}$
$$
\mathrm{E}=\frac{1}{4 \varepsilon_0} \mathrm{Kr}^2
$$
So, direction of $\mathrm{E}$ is radially outwards.
Case II : By Gauss's law
$$
\oint E . d S=\frac{q}{\epsilon_0}
$$
For points $r>R$, electric field intensity will be
$$
\oint \mathrm{E} \cdot \mathrm{dS}=\frac{1}{\varepsilon_0} \int \rho \mathrm{dV} \quad(\because \mathrm{q}=\rho \mathrm{dV})
$$
FordV,
$$
\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3
$$
$$
\begin{aligned}
\mathrm{dV} &=3 \times \frac{4}{3} \pi \mathrm{r}^2 \mathrm{dr} \\
&=4 \pi \mathrm{r}^2 \mathrm{dr} \\
\rho(\mathrm{r}) &=\mathrm{Kr} \text { for } \mathrm{r}>\mathrm{R} \\
\mathrm{E}\left(4 \pi \mathrm{r}^2\right)=\frac{4 \pi \mathrm{K}}{\varepsilon_0} \int_0^{\mathrm{R}} \mathrm{r}^3 \mathrm{dr}=\frac{4 \pi \mathrm{K}\left(\mathrm{R}^4-0\right)}{\varepsilon_0} 4
\end{aligned}
$$
$\mathrm{E}=\frac{\mathrm{K}}{4 \varepsilon_0} \frac{\mathrm{R}^4}{\mathrm{r}^2}$
Charge density is again positive. So, the direction of $E$ is radially outward.
(b) The two protons must be on the opposite sides of the centre along a diameter following, the rule of symmetry. This can be shown by the figure so, total negative charge on the sphere is $2 \mathrm{e}$, it is distributed in sphere of radius $R$.

Symmetrically,
so, the charge on sphere is
$$
\begin{aligned}
\mathrm{q} &=\int_0^{\mathrm{R}} \rho \mathrm{dV}=\int_0^{\mathrm{R}}(\mathrm{Kr}) 4 \pi \mathrm{r}^2 \mathrm{dr} \\
\mathrm{q} &=4 \pi \mathrm{K} \frac{\mathrm{R}^4}{4}=2 \mathrm{e} \\
\therefore \quad \mathrm{K} &=\frac{2 \mathrm{e}}{\pi \mathrm{R}^4}
\end{aligned}
$$
If protons 1 and 2 are embedded at distance $r$ from the centre $\mathrm{O}$ of the sphere as shown thus attractive force on proton 1 due to charge distribution inside the charge sphere at $\mathrm{r} < \mathrm{R}$ from part (a) is
$$
F_1=\mathrm{eE}=\frac{-\mathrm{eKr}^2}{4 \varepsilon_0}, \quad\left(\because \mathrm{E}=\frac{\mathrm{Kr}^2}{4 \epsilon_0}\right)
$$
Repulsive force on proton 1 due to proton 2 is
$$
\mathrm{F}_2=\frac{\mathrm{e}^2}{4 \pi \varepsilon_0(2 \mathrm{r})^2}
$$
Net force on proton 1 ,
$$
\begin{aligned}
&\mathrm{F}=\mathrm{F}_1+\mathrm{F}_2 \\
&\mathrm{~F}=\frac{-\mathrm{eKr}^2}{4 \varepsilon_0}+\frac{\mathrm{e}^2}{16 \pi \varepsilon_0 \mathrm{r}^2}
\end{aligned}
$$
By putting the value of $(\mathrm{K})$
So,
$$
\mathrm{F}=\left[\frac{-\mathrm{er}^2}{4 \varepsilon_0} \frac{2 \mathrm{e}}{\pi \mathrm{R}^4}+\frac{\mathrm{e}^2}{16 \pi \varepsilon_0 \mathrm{r}^2}\right]
$$
Therefore, net force on proton 1 will be zero, so
$$
\begin{aligned}
\mathrm{F}=0 \quad \text { or } \quad \mathrm{F}_1+\mathrm{F}_2=0 \\
\frac{-\mathrm{er}^2 2 \mathrm{e}}{4 \epsilon_0 \pi \mathrm{R}^4}+\frac{\mathrm{e}^2}{16 \pi \epsilon_0 \mathrm{r}^2}=0 \\
\frac{\mathrm{er}^2 2 \mathrm{e}}{4 \varepsilon_0 \pi \mathrm{R}^4}=\frac{\mathrm{e}^2}{16 \pi \varepsilon_0 \mathrm{r}^2} \\
\mathrm{r}^4=\frac{\mathrm{R}^4}{8} \quad \text { or } \quad \mathrm{r}=\frac{\mathrm{R}}{(8)^{1 / 4}}
\end{aligned}
$$
This is the distance of each of the two protons from the centre of the sphere.