$6\mathit{\mu }\mathrm{F}$ and $3\mathit{\mu }\mathrm{F}$ are connected in series. Hence their equivalent capacitance is $2\mathit{\mu }\mathrm{F}$.
$4\mathit{\mu }\mathrm{F}$ and $2\mathit{\mu }\mathrm{F}$ are connected in series. Hence their equivalent capacitance is $\frac{4}{3}\mathit{\mu }\mathrm{F}$.
$2\mathit{\mu }\mathrm{F}$ and $\frac{4}{3}\mathit{\mu }\mathrm{F}$ capacitors are connected in parallel.
So, the equivalent capacitance $=\frac{10}{3}\mathit{\mu }\mathrm{F}=\frac{10}{3}\times {10}^{-6}\mathrm{F}$
$\mathrm{Q}=\mathrm{CV}$
$\mathrm{C}=\frac{10}{3}\times {10}^{-6}\mathrm{F}$
$\mathrm{V}=3\mathrm{V}$
$\therefore \mathrm{}\mathrm{Q}=\frac{10}{3}\times {10}^{-6}\times 3={10}^{-5}\mathrm{C}$