A positively charged sphere of radius $r_0$ carries a volume charge density $\rho$ (Figure). A spherical cavity of radius  $r_0/2$ is then scooped out and left empty, as shown. What is the direction and magnitude of the electric field at point B?

This sphere is an equivalent to the superposition of a big sphere of radius r₀ with a charge density of $\rho$ and a smaller sphere of radius $\frac{r_0}{2}$ with a charge density of $-\rho$

Electric field on surface of a uniformly charged sphere is given by $\frac{Q}{4\pi {\epsilon }_0{R}^3}=\frac{\rho R}{3{\epsilon }_0}$

Electric field at outside point is given by

$E=\frac{Q}{4\pi {\epsilon }_0{r}^2}=\frac{\rho R^3}{3{\epsilon }_0{r}^2}$

The electric field at 'B' will be the superposition of the field due to the larger sphere and smaller sphere.

$\left|\overline{E_r}\right|=\frac{\rho r_0}{3{\epsilon }_0}-\frac{\rho {\left(\frac{r_0}{2}\right)}^3}{3{\epsilon }_0{\left(\frac{3r_0}{2}\right)}^2}=\frac{17\rho r_0}{54{\epsilon }_0}left$