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 $$\frac{|\vec{E_A}|}{|\vec{E_B}|}$$ of magnitude of electric field $$\vec{E_A}$$ and $$\vec{E_B}$$, respectively, at points $$A$$ and $$B$$ due to the remaining portion is:

Electric field of a uniformly charged full sphere at an internal point $$r \le R$$ and external point $$r > R$$:

$$E_{\text{in}} = \frac{\rho r}{3\varepsilon_0}$$

$$E_{\text{out}} = \frac{\rho R^3}{3\varepsilon_0 r^2}$$

For point A (center of the large sphere, surface of the cavity):

$$E_{\text{full, A}} = 0$$

$$E_{\text{cavity, A}} = \frac{\rho (R/2)}{3\varepsilon_0} = \frac{\rho R}{6\varepsilon_0}$$

$$E_A = |E_{\text{full, A}} - E_{\text{cavity, A}}| = \frac{\rho R}{6\varepsilon_0}$$

For point B (bottom surface of the large sphere):

$$E_{\text{full, B}} = \frac{\rho R}{3\varepsilon_0}$$

$$E_{\text{cavity, B}} = \frac{\rho (R/2)^3}{3\varepsilon_0 (3R/2)^2} = \frac{\rho R^3}{24\varepsilon_0 \cdot \frac{9R^2}{4}} = \frac{\rho R}{54\varepsilon_0}$$

$$E_B = E_{\text{full, B}} - E_{\text{cavity, B}} = \frac{\rho R}{3\varepsilon_0} - \frac{\rho R}{54\varepsilon_0} = \frac{17\rho R}{54\varepsilon_0}$$

$$\frac{|E_A|}{|E_B|} = \frac{\frac{\rho R}{6\varepsilon_0}}{\frac{17\rho R}{54\varepsilon_0}} = \frac{1}{6} \times \frac{54}{17} = \frac{9}{17} = \frac{18}{34}$$