Consider  a sphere of radius R which carries a uniform charge density  $\rho$.  If  a sphere of radius $\frac{R}{2}$  is  carved out of  it, as shown , the ratio $\frac{\left|{\overrightarrow{E}}_A\right|}{\left|{\overrightarrow{E}}_B\right|}$of magnitude of electric field ${\overrightarrow{E}}_A$and  ${\overrightarrow{E}}_B$ . respectively, at points A and B due to the  remaining portion is:

$\begin{array}{l}Assume charge density inside cavity and outside cavity is +\rho and -\rho respectively.\text{}\\ E_{in}=\frac{\rho r}{3{\in }_0} \left(r<R\right) \text{} \\ E_{out}=\left(\frac{\rho R^3}{3{\in }_0}\right)\frac{1}{r^2} \left(r>R\right)\\ Point A is inside both spheres, \text{}\\ E_A =E_1+E_2 = \frac{\rho }{3{\epsilon }_0}\left(0\right)+\frac{-\rho }{3{\epsilon }_0}\left(\frac{R}{2}\right) = \frac{-\rho R}{6{\epsilon }_0}\\ Point B inside for \left(1\right) and outside for \left(2\right)\text{}\\ E_B =E_1+E_2 = \frac{\rho R}{3{\epsilon }_0}+\frac{-\rho }{3{\epsilon }_0}\frac{{\left(\frac{R}{2}\right)}^3}{{\left(\frac{3R}{2}\right)}^2}=\frac{17\rho R}{54{\epsilon }_0}\\ Ratio = \frac{E_A }{E_B }=\frac{18}{34}\end{array}$