In the combination, the capacitors 6 $\mathrm{\mu }$F,4 $\mathrm{\mu }$F and 8 $\mathrm{\mu }$F are in parallel. Their capacitance is 18 $\mathrm{\mu }$F. There arc two other capacitors of capacitance 18 $\mathrm{\mu }$F. Now these capacitors are in series. Hence, total capacitance is $18/3=6\mathrm{\mu F}$.
$\therefore$Potential drop across each of the 6 $\mathrm{\mu }$F, 4 $\mathrm{\mu }$F and 8 $\mathrm{\mu }$F capacitors is 10 V.
So, charge g across 4 $\mathrm{\mu }$F capacitor $=10\times 4=40 \mathrm{\mu C}$