Electrostatics Part 1

Use superposition.The remaining body is the complete sphere of density $\rho$ minus the removed cavity sphere of density $\rho$.Thus $\vec E_{\text{remaining}}=\vec E_{\text{full}}-\vec E_{\text{cavity}}$.Let $O$ be the centre of the large sphere.From the figure, the cavity has centre $C=(0,R/2)$ and radius $R/2$.Point $A$ is the right end of the horizontal radius of the cavity, so $A=(R/2,R/2)$.Point $B$ is the bottom point of the large sphere, so $B=(0,-R)$.For the complete sphere, inside it $\vec E_{\text{full}}=\frac{\rho}{3\epsilon_0}\vec r$.At $A$: $\vec E_{\text{full},A}=\frac{\rho}{3\epsilon_0}\left(\frac{R}{2}\hat i+\frac{R}{2}\hat j\right)=\frac{\rho R}{6\epsilon_0}(\hat i+\hat j)$.At the surface of the cavity, $\vec E_{\text{cav},A}=\frac{\rho (R/2)}{3\epsilon_0}\hat i=\frac{\rho R}{6\epsilon_0}\hat i$.So $\vec E_A=\vec E_{\text{full},A}-\vec E_{\text{cav},A}=\frac{\rho R}{6\epsilon_0}\hat j$.Therefore $|\vec E_A|=\frac{\rho R}{6\epsilon_0}$.At $B$, the cavity behaves as a point charge $Q_{\text{cav}}=\rho\frac{4}{3}\pi\left(\frac{R}{2}\right)^3=\frac{\rho\pi R^3}{6}$.The full sphere gives $|\vec E_{\text{full},B}|=\frac{\rho R}{3\epsilon_0}$ downward.The cavity gives $|\vec E_{\text{cav},B}|=\frac{1}{4\pi\epsilon_0}\frac{Q_{\text{cav}}}{(3R/2)^2}=\frac{\rho R}{54\epsilon_0}$ downward.Thus $|\vec E_B|=\frac{\rho R}{3\epsilon_0}-\frac{\rho R}{54\epsilon_0}=\frac{17\rho R}{54\epsilon_0}$.Hence $\left|\frac{\vec E_A}{\vec E_B}\right|=\frac{\rho R/(6\epsilon_0)}{17\rho R/(54\epsilon_0)}=\frac{9}{17}=\frac{18}{34}$.So the correct option is $\frac{18}{34}$.