The charge on the $4\mathrm{\mu F}$ capacitor in the steady state is

In steady state, the capacitors are fully charged and act as open circuit, so the equivalent circuit in steady state would be as shown:-

$\therefore$ Steady State current is $I=\frac{12}{2+4}=2 \mathrm{A}$
$\therefore$ Potential difference across $\mathrm{AB}$ is $\mathrm{V}=2\times 4=8\mathrm{ }\mathrm{V}$
$\therefore$ Sum of potential difference across $2 \mathrm{\mu F}$ and $4 \mathrm{\mu F}$capacitors is $8 \mathrm{V}$. As these are in series, so charges on them would be same. Let $q$ be the charge on them, then:-
From KVL, 

$$\begin{gathered} \frac{q}{2}+\frac{q}{4}=8 \\ \Rightarrow q=\frac{32}{3} \mathrm{\mu C} \end{gathered}$$