Explanation

(c) When the switch is open, $3\mu F$ and $6\mu F$ capacitors are in series. Hence charge on each capacitor
$q = C_{eq}V = \frac{3 \times 6}{3 + 6} \times 9 = 18\ \mu C$
When the switch is closed, in the steady state no current will flow through the capacitor. Therefore the two resistors $3\Omega$ and $6\Omega$ will be in series. Current in each resistor will be
$I = \frac{9}{3 + 6} = 1\ A$
Now the $3\mu F$ capacitor and $6\mu F$ capacitor will be in parallel with $3\Omega$ and $6\Omega$ resistor respectively
Charge on $3\mu F$ capacitor $q_{1} = CV = 3 \times 3 = 9\mu C$
Charge on $6\mu F$ $q_{2} = CV = 6 \times 6 = 36\mu C$
Charge flowing through the switch = increase in charge on the system consisting of right plate of $3\mu F$ and left plate of $6\mu F = ( - 9 + 36) = 27\mu C$