Let the capacitors be `C_(1)=8muF,C_(2)=8muFandC_(3)=4muF`.

All the capacitors are connected in series, so, the equivalent capacitance `C_(s)` can be calculated as

`(1)/(C_(s))=(1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3))`

`(1)/(C_(s))=(1)/(8)+(1)/(8)+(1)/(4)`

`(1)/(C_(s))=(1+1+2)/(8)`

`C_(s)=2muF`

In series combination,

`Q_(1)=Q_(2)=Q_(3)`

But `Q=C_(s)V`

`:.` Charge on capacitor of capacity `4muF` can be calculated as

`Q=2xx10^(-6)xx120`

`=240xx10^(-6)C=240muC`