The solution to the problems is as follows:
(i) Charge on each capacitors:
The capacitors C1 and C2 are in series. Therefore the equivalent capacitance of this combination is
As you know, charge Q= CV.
Therefore charge on C1= charge on C2=3µF×12V=36µC.
Charge on C3=6µF×12V=72µC.
(ii) Equivalent capacitance of the network:
The equivalent capacitance of the series combination of C1 and C2 is 3µF. This is in parallel with C3. Therefore, the equivalent capacitance of the network Ceq=3µF+ 6µF =9µF.
(iii) Energy stored in the network of capacitors: