Electric Flux & Gauss's Law question from JEE Main 2020 (9 Jan, Shift 1)

Answer: (c)

1. Treat the cavity as a full sphere of $+\rho$ plus a small sphere of $-\rho$ (radius $\frac{R}{2}$, centre $\frac{R}{2}$ above O). A is at O, B is at distance $R$ below O.
2. At A: big sphere gives 0 (centre); small sphere, A on its surface: $\frac{\rho (R/2)}{3\varepsilon_0}$. So $E_A = \frac{\rho R}{6\varepsilon_0}$.
3. At B: big sphere gives $\frac{\rho R}{3\varepsilon_0}$; small sphere, B at $\frac{3R}{2}$ from its centre: $\frac{\rho (R/2)^3}{3\varepsilon_0 (3R/2)^2} = \frac{\rho R}{54\varepsilon_0}$, opposite in direction.
4. $E_B = \frac{\rho R}{3\varepsilon_0} - \frac{\rho R}{54\varepsilon_0} = \frac{17\rho R}{54\varepsilon_0}$
5. $\frac{E_A}{E_B} = \frac{1/6}{17/54} = \frac{9}{17} = \frac{18}{34}$
6. So the answer is (c) $\frac{18}{34}$.