In the circuit shown in the figure, the total charge is $750\mu C$ and the voltage across capacitor $C_2$ is 20 V.  Then the charge on capacitor $C_2$ is  $\left(in\mu C\right)$

According to question,

$Q=750\mu C=q_2+q_3$

Capacitors  $C_{2}and{C}_3$ are in parallel hence,

Voltage across  $C_2=$voltage across $C_3=20V$

Change on capacitor $C_3$, 

$q_3=C_3\times V_3=8\times 20=160\mu C$

$\therefore q_2=750\mu C-160\mu C=590\mu C$