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