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

Electric field at  $A(R\text{'}=\frac{R}{2})$
$$\begin{gathered} E_A.ds=\frac{q}{{\epsilon }_0} \\ \Rightarrow {\overrightarrow{E}}_A=\frac{\rho \times \frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3}{{\epsilon }_0.4\pi {\left(\frac{R}{2}\right)}^2} \\ \Rightarrow {\overrightarrow{E}}_A=\frac{\sigma (R/2)}{3{\epsilon }_0}=\left(\frac{\sigma R}{6{\epsilon }_0}\right) \end{gathered}$$ 
Electric fields at ‘B’
$$\begin{gathered} {\overrightarrow{E}}_B=\frac{k\times \rho \times \frac{4}{3}\pi R^2}{R^2}-\frac{k\times \rho \times \frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3}{{\left(\frac{3R}{2}\right)}^2} \\ \Rightarrow {\overrightarrow{E}}_B=\frac{\sigma R}{3{\epsilon }_0}-\left(\frac{1}{4\pi {\epsilon }_0}\right)\frac{\left(\sigma \right)}{{\left(\frac{3R}{2}\right)}^2}\frac{4\pi }{3}{\left(\frac{R}{2}\right)}^3 \\ \Rightarrow {\overrightarrow{E}}_B=\frac{\sigma R}{3{\epsilon }_0}-\frac{\sigma R}{54{\epsilon }_0}\Rightarrow E_B=\frac{17}{54}\left(\frac{\sigma R}{{\epsilon }_0}\right) \\ \left|\frac{E_A}{E_B}\right|=\frac{1\times 54}{6\times 17}=\left(\frac{9}{17}\right)=\frac{9}{17}\times \frac{2}{2}=\frac{18}{34} \end{gathered}$$