Answer
$0.61mJ;0.18mJ;65.68\mu J$
Work Step by Step
The energy stored in each capacitor can be determined as
$U_1=\frac{1}{2}CV^2$
$\implies U_1=\frac{1}{2}(15\times 10^{-6}F)(9.0V)^2$
$U_1=0.61mJ$
$C_{eq}=\frac{(8.2\times 10^{-6}F)(22\times 10^{-6}F)}{8.2\times 10^{-6}F+22\times 10^{-6}F}=5.97\times 10^{-6}F$
$Q=C_{eq}V$
$Q=(5.97\times 10^{-6}F)(9.0V)=53.76\times 10^{-6}C$
Now $U_2=\frac{Q^2}{2C_2}$
$U_2=\frac{(53.76\times 10^{-6}F)^2}{2(8.2\times 10^{-6}F)}=0.18mJ$
and $U_3=\frac{Q^2}{2C_3}$
$\implies U_3=\frac{(53.76\times 10^{-6}C)^2}{2(22\times 10^{-6}F)}$
$\implies U_3=65.68\mu J$
