Let $$P(r) = \frac{Q}{\pi R^4}r$$ be the charge density distribution for a solid sphere of radius $$R$$ and total charge $$Q$$. For a point '$$p$$' inside the sphere at distance $$r_1$$ from the centre of the sphere, the magnitude of electric field is

The charge density varies with the distance from the centre as
$$\rho(r)=\frac{Q}{\pi R^{4}}\;r \qquad (0\le r\le R)$$

Step 1: Verify that the given $$\rho(r)$$ produces total charge $$Q$$.
Total charge inside the full sphere is
$$Q_{\text{total}}=\int_{0}^{R}\rho(r)\,dV =\int_{0}^{R}\rho(r)\,4\pi r^{2}\,dr$$
$$=4\pi\int_{0}^{R}\frac{Q}{\pi R^{4}}\,r\,(r^{2})\,dr =\frac{4Q}{R^{4}}\int_{0}^{R}r^{3}\,dr =\frac{4Q}{R^{4}}\Bigl[\frac{r^{4}}{4}\Bigr]_{0}^{R}=Q.$$
Thus the distribution is consistent.

Step 2: Charge enclosed by a Gaussian sphere of radius $$r_{1}\;(r_{1}\lt R)$$.
$$q_{\text{enc}}=\int_{0}^{r_{1}}\rho(r)\,4\pi r^{2}\,dr =4\pi\int_{0}^{r_{1}}\frac{Q}{\pi R^{4}}\,r^{3}\,dr$$
$$=\frac{4Q}{R^{4}}\Bigl[\frac{r^{4}}{4}\Bigr]_{0}^{r_{1}} =\frac{Q\,r_{1}^{4}}{R^{4}}.$$

Step 3: Apply Gauss’s law to find the electric field magnitude at $$r_{1}$$.
For a spherical Gaussian surface,
$$E(4\pi r_{1}^{2})=\frac{q_{\text{enc}}}{\varepsilon_{0}} \;\;\Longrightarrow\;\; E=\frac{q_{\text{enc}}}{4\pi\varepsilon_{0}r_{1}^{2}}$$
Insert the value of $$q_{\text{enc}}$$:
$$E=\frac{1}{4\pi\varepsilon_{0}r_{1}^{2}}\;\frac{Q\,r_{1}^{4}}{R^{4}} =\frac{Q\,r_{1}^{2}}{4\pi\varepsilon_{0}R^{4}}.$$

Hence, the magnitude of the electric field at distance $$r_{1}$$ inside the sphere is
$$\boxed{\dfrac{Q\,r_{1}^{2}}{4\pi\varepsilon_{0}R^{4}}}.$$

Option C which is: $$\frac{Qr_{1}^{2}}{4\pi\varepsilon_{0}R^{4}}$$