In the given circuit the potential difference across the $7\mu F$ capacitor is $6V$. Then 

Since same charge flows through $7\mu F$ and $3\mu F$ (as both are in series), so

$$\begin{gathered} Q_7=Q_3=42\mu F \\ \Rightarrow 7\times 6=3V_3 \\ {V}_3=14V \end{gathered}$$

So $V_{3.9}=20V$

Charge on $12\mu F$ capacitor is

$$\begin{gathered} Q_{12}=Q_{3.9}+Q_7 \\ {Q}_{12}=3.9\times 20+\frac{7\times 3}{7+30}\times 20 \end{gathered}$$

$Q_{12}=120\mu C$ $=$Total charge in the circuit

$V_{12}=\frac{120}{12}=10V$

Voltage across battery is $14+6+10=30V$