RC circuit

At steady state, no current passes through the capacitors. For ease, we can just remove them from the circuit. After distributing the current with help of Kirchhoff's laws, the circuit is like the one above. It is clear that the potential across the points $A$ and $B$ is the potential across the capacitor ${C}_{2}$ .
${V}_{A}+10(\frac{8}{9})-10(\frac{10}{9})-10(\frac{2}{9})={V}_{B}\\{V}_{A}-{V}_{B}=\frac{40}{9}\\Q=9(\frac{40}{9})\\\Huge\color{#3D99F6}{=\boxed{40 \mu C}}$

In the steady state we know that any current that pass through capacitor is equal to zero.

So, we reduce our problem from DE to simple linear equation. We have

$2=2i_{1}+i_{2}$

$2=i_{1}+5i_{2}$

From this we have $i_{2}=\frac{2}{9} A$

then we got $Q=9*\frac{40}{9}=40 \mu C$