charge $Q=C_{e q} V=4 V \mu C$

The charge on the $7 \mu F$ or $3 \mu F$ capacitor is

$Q_{2}=(7 \mu F)(6 V)=42 \mu C$

$\frac{Q_{2}}{2.1 \mu F}=\frac{Q_{1}}{12 \mu F}$ or $Q_{1}=(42 \mu C) \frac{(3.9 \mu F)}{(2.1 \mu F)}=78 \mu C$

$Q=Q_{1}+Q_{2}=(78 \mu C+42 \mu C)=120 \mu C=4 V \mu C$

Emf of the battery is $V=30 V$

The potential drop across $12 \mu F$ capacitor is

$\frac{Q}{12 \mu F}=\frac{120 \mu C}{12 \mu F}=10 V$