In the given circuit, if the potential difference between $A$ and $B$ is $80 \mathrm{V},$ then the equivalent capacitance between $A$ and $B$, and the charge on $10 \mu \mathrm{F}$ capacitor respectively, are

Simplifying the circuit as in the diagram, 

Capacitors $10 \mathrm{\mu F}, 5 \mathrm{\mu F} \text{and} 9 \mathrm{\mu F}$are in parallel therefore, net capacitance across the point $C\text{ and }D$is, $C_{CD}=10+5+9=24 \mathrm{\mu F}$

Now, capacitors $12 \mathrm{\mu F}, 24 \mathrm{\mu F} \text{and} 8 \mathrm{\mu F}$are in series connection 

Equivalent capacitor $\frac{1}{C_{eq}}=\frac{1}{12}+\frac{1}{24}+\frac{1}{8}=\frac{1}{4}$

$\Rightarrow C_{eq}=4 \mathrm{\mu F}$

For the charge on the capacitor $10 \mathrm{\mu F},$

The net charge flowing in the circuit, $Q=C{V}_{AB}=4 \mathrm{\mu F}\times 80 \mathrm{V}=320 \mathrm{\mu C}$

Potential difference across $CD$is, $V_{CD}=\frac{Q}{C_{CD}}=\frac{320}{24}=\frac{40}{3} \mathrm{V}$

Therefore, charge on the $10 \mathrm{\mu F}$capacitor is $q=10 \mathrm{\mu F}\times V_{CD}=10 \mathrm{\mu F}\times \frac{40}{3} \mathrm{V}=133 \mathrm{\mu C}$