b
When $S$ is closed, the energy stored in the capacitor will be dissipated through resistor.
Thus, $\frac{Q^{2}}{2 C}=3.6 \times 10^{-3}$
$Q^{2}=3.6 \times 10^{-3} \times 2 \times 2 \times 10^{-6}=144 \times 10^{-10}$
$Q=12 \times 10^{-5} C=120 \mu C$