In the circuit in figure, the ratio of charges on  $5\mu F$ and $4\mu F$ capacitor is:

Charge on $4\mu F$ capacitor is $Q=4\times 6=24\mu C$ . Equivalent of $2\mu Fand5\mu F$in parallel is $7\mu F$ . This $7\mu F$  in series with $3\mu F$  makes  $\frac{7\times 3}{10}=\frac{21}{10}\mu F.$ Charge on the equivalent capacitance is 
$q=\frac{21}{10}\times 6=\frac{63}{5}\mu C$
This charge q gets divided on $2\mu Fand5\mu F$  so that potential difference across them same. It means charges on them are in ratio of their capacitances.
∴ Charge on 5mF capacitance is  $\frac{q\times 5}{5+2}$
$=\frac{63}{5}\times \frac{5}{7}=9\mu C$
Required ratio is  $\frac{9\mu C}{24\mu C}=\frac{3}{8}$