Answer : (a) $\;\large\frac{\rho_{0}}{\in_{0}}\;[\large\frac{r}{3}-\large\frac{r^2}{4R}]$

Explanation :

Due to symmetry , if electric field exists then it will be radial

Let the electric field inside the sphere at a distance r from the centre is $\;E_{r}$

Then flux through a shell of radius r is :

By Gauss 's law

$E_{r}\;.4 \pi r^2=\phi$

$\phi=E_{r} 4 \pi r^2$

$\phi=\large\frac{q_{inc}}{\in_{0}}$

$E_{r} 4 \pi r^2=\large\frac{q_{inc}}{\in_{0}}---(1)$

$q_{inc}\;$= charge enclosed in the shell of radius r

$q_{inc}= \int \; \rho dV$

Consider a shell of radius r and thichness dr

$q_{inc}= \int _{0}^{r}\;\rho_{0}\;(1-\large\frac{r}{\rho})\;4 \pi r^2 dr$

$q_{inc}=4 \pi \rho_{0}\; \int_{0}^{r}\;(r^2-\large\frac{r^3}{R}) dr$

$=4 \pi \rho_{0} \;(\large\frac{r^3}{3}-\large\frac{r^4}{4R})$

Putting the value of $\;q_{inc}\;$ in equation (1) we get

$E_{r} 4 \pi r^2=\large\frac{4 \pi \rho_{0}}{\in_{0}}\;(\large\frac{r^3}{3}-\large\frac{r^4}{4R})$

$E_{r}=\large\frac{\rho_{0}}{\in_{0}}\;(\large\frac{r}{3}-\large\frac{r^2}{4R})$