A solid ball of radius R has a charge density $\rho$ given by $\mathrm{\rho }={\mathrm{\rho }}_0\left(1-\frac{\mathrm{r}}{\mathrm{R}}\right)\text{ for }0\le \mathrm{r}\le \mathrm{R}$The electric field outside the ball is for r >R :

Charge density,
$\begin{array}{l}\mathrm{\rho }={\mathrm{\rho }}_0\left(1-\frac{\mathrm{r}}{\mathrm{R}}\right)\\ \mathrm{dq}=\mathrm{\rho dv}\\ {\mathrm{q}}_1=\int \mathrm{dq}=\mathrm{\rho dv}\\ ={\mathrm{\rho }}_0\left(1-\frac{\mathrm{r}}{\mathrm{R}}\right)4{\mathrm{\pi r}}^2\mathrm{dr}\left(\therefore \mathrm{dv}=4{\mathrm{\pi r}}^2\mathrm{dr}\right)\\ =4{\mathrm{\pi \rho }}_0{\int }_0^{\mathrm{k}} \left(1-\frac{\mathrm{r}}{\mathrm{R}}\right){\mathrm{r}}^2\mathrm{dr}=4{\mathrm{\pi \rho }}_0{\int }_0^{\mathrm{k}} {\mathrm{r}}^2\mathrm{dr}-\frac{{\mathrm{r}}^2}{\mathrm{R}}\mathrm{dr}\\ =4{\mathrm{\pi \rho }}_0\left[{\left[\frac{{\mathrm{r}}^3}{3}\right]}_0^{\mathrm{R}}-{\left[\frac{{\mathrm{r}}^4}{4\mathrm{R}}\right]}_0^{\mathrm{R}}\right]=4{\mathrm{\pi \rho }}_0\left[\frac{{\mathrm{R}}^3}{3}-\frac{{\mathrm{R}}^4}{4\mathrm{R}}\right]\\ =4{\mathrm{\pi \rho }}_0\left[\frac{{\mathrm{R}}^3}{3}-\frac{{\mathrm{R}}^4}{4}\right]=4{\mathrm{\pi \rho }}_0\left[\frac{{\mathrm{R}}^3}{12}\right]\\ \mathrm{q}=\frac{{\mathrm{\pi \rho }}_0{\mathrm{R}}^3}{3}\\ \mathrm{E}. 4{\mathrm{\pi r}}^2=\left(\frac{{\mathrm{\pi \rho }}_0{\mathrm{R}}^3}{3{\mathrm{ϵ}}_0}\right)\end{array}$
Electric field outside the ball ,$\mathrm{E}=\frac{{\mathrm{\rho }}_0{\mathrm{R}}^3}{12{\in }_0{\mathrm{r}}^2}$