Solution :
Using the Gauss law, the electric field intensity of a sphere is given by :
[tex]$E. 4 \pi r^2 = \frac{Q_{enc}}{\epsilon_0}$[/tex]
[tex]$E = \frac{Q_{enc}}{4 \pi r^2 \epsilon_0}$[/tex]
Now the enclosed charge inside the sphere (r<a) is
[tex]$Q_{enc}= \rho V$[/tex]
[tex]$Q_{enc}= \rho \left( \frac{4}{3} \pi r^3 \right)$[/tex]
Hence, the electric field intensity becomes as follows :
[tex]$E= \frac{\rho \left(\frac{4}{3} \pi r^3\right)}{4 \pi \epsilon_0 r^2}$[/tex]
[tex]$E =\frac{\rho r}{3 \epsilon_0}$[/tex]
Thus, the electric field inside the sphere is given by :
[tex]$\vec {E} = \frac{\rho \vec{r}}{3 \epsilon_0}$[/tex]
