Hint: To find the effective capacitance you probably used the following steps:
(i) Replace the $5 \mu F$ and $6 \mu F$ capacitors by a single effective capacitor, say $C_1$;
(ii) Replace the $2 \mu F$ and $4 \mu F$ capacitors by another single effective capacitor, say $C_2$;
(iii) You now have two capacitors in parallel and you can now replace the original network by a single effective capacitor, say $C$.
(iv) Now retrace your steps: The capacitor in step (iii), $C$, has a 90 V battery connected across it. But this came from the workings in step (ii). This means that the p.d. across each of the terminals of each of the two (parallel) capacitors $C_1, C_2$ is 90 V so that you can calculate WHAT for each (effective) capacitor?
(v) Now go back one further step, each of the capacitors $C_1, C_2$ above is really a pair of capacitors in series. Just look at, for example, the top pair of series capacitors that you used to determine $C_1$. You know the charge on each of the capacitors since for capacitors in series are the same (as charge accumulates on the left hand plate of the $5 \mu F$ capacitor, an equal amount of charge is repelled on the right hand plate, this flows onto the left hand plate of the $6 \mu F$ plate, that repels an equal amount of charge on the right hand plate). This amount of charge was calculated in the step (iv).