The capacitors are connected in series then the net capacitance will be, $C_{net}=\frac{C_1\times C_2}{C_1+C_2}=\frac{2\times 8}{2+8}=1.6 \mathrm{mF}$, then the net charge will be, $Q_{net}=C_{net}V=1.6\times {10}^{-3}\times 300=480\times {10}^{-3} \mathrm{C}$, which will be same in each capacitor now potential difference across each capacitor will be, 
$V_1=\frac{Q_{net}}{C_1}=\frac{480\times {10}^{-3} \mathrm{C}}{2\times {10}^{-3}}=240 \mathrm{V}$ and $V_2=\frac{Q_{net}}{C_2}=\frac{480\times {10}^{-3} \mathrm{C}}{8\times {10}^{-3}}=60 \mathrm{V}$

and the energy stored will be, 
$U=\frac{1}{2}{C}_{net}{\left(V\right)}^2=\frac{1}{2}\times 1.8\times {10}^{-3}\times {\left(300\right)}^2=81 \mathrm{J}$