Combination of Capacitors question from JEE Main 2016

Answer: (a)

1. $3 \parallel 9 = 12\ \mu\text{F}$, in series with $4$: $3\ \mu\text{F}$, across 8 V: $Q = 24\ \mu\text{C}$ (on the $4\ \mu\text{F}$).
2. Voltage on the parallel pair: $\frac{24}{12} = 2$ V, so $Q_9 = 18\ \mu\text{C}$.
3. $Q = 24 + 18 = 42\ \mu\text{C}$
4. $E = \frac{kQ}{r^2} = \frac{9\times10^9 \times 42\times10^{-6}}{900} = 420$ N/C
5. So the answer is (a) 420 N/C.