A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is
 

Initial charge on capacitor 
 $q=C_{eff}V$
 $q=\frac{3\times 6}{3+6}\times 9=18\mu C$
When switch S is closed
At steady state  $i=\frac{V}{R_{eff}}=\frac{9}{9}=1A$
$3\Omega$ Resistor &  $3\mu F$ capacitor are in parallel similarly  $6\Omega$ &  $6\mu F$ are in parallel
 $\therefore V_{\left(3\Omega \right)}=V_{\left(3\mu F\right)} V_{\left(6\Omega \right)}=V_{\left(6\mu F\right)}$   
$1\times 3=V_{\left(3\mu F\right)} 1\times 6=V_{\left(6\mu F\right)}$    
Final charges on capacitor
$q_{\left(3\mu F\right)}=3\times 3=9\mu C q_{\left(6\mu F\right)}=6\times 6=36\mu C$