Two capacitors having capacitances $8\mu Fand16\mu F$  have breaking voltages 20 V and 80 V. They are combined in series. The maximum energy we can store in the combination is 

Maximum charge on $8\mu F$  capacitor that is allowed is  $Q_1=8\times 20=160\mu F$.

Maximum charge on  $16\mu F$ capacitor can be  $Q_2=16\times 80=1280\mu C$. 

When in series, the maximum charge on both of them shell be $160\mu C$  so that none of the capacitor punctures.

$\therefore Q_{\max }=160\mu C$
$\therefore U_{\max }=\frac{{\left(160\right)}^2}{2\times 8}+\frac{{\left(160\right)}^2}{2\times 16}=2400\mu J$