In the given circuit , the charge on 4$\mathrm{\mu F}$ capacitor will be

Equivalent simplified circuit will be $C_{eq}=(3+\frac{4\times 6}{4+6})=5.4\mu F$

Total charge $q=C_{eq}.V=5.4\times 10=54\mu C$

Charge distribution in the circuit will be as 

$\frac{q_1}{q_2}=\frac{2.4}{3}=\frac{4}{5}=k,q_1=4K,q_2=5k$

Total charge , $9k=54\mu c\Rightarrow k=6\mu c$

Charge on $4\mu F\&6\mu F$will be same 

$=4\times 6\mu c=24\mu c$