In the circuit shown in Figure, 7 µC of charge is given to the right plate of the 5 µF capacitor. Then, in the steady state, the charge on the left place of the 4 µF capacitor will be

So the charge acquired on the right plate of $5\mu F$ capacitor will equal to some of the charges on the 4µF and 3µF.

Let q µC be the charge on the left plate of the 4 µF capacitor. Then the charge on the left plate of the 3 µF capacitor will be (7$-$q) µC. Since the potential difference across the 3 µF capacitor = potential difference across the 4 µF capacitor,

$\begin{array}{rcl}\frac{7-q}{3}& =& \frac{q}{4}\\ q& =& 4.0 \mu C\end{array}$