INITIAL energy stored in capacitor \[2\mu F\] \[{{U}_{i}}=\frac{1}{2}2{{(V)}^{2}}={{V}^{2}}\] FINAL VOLTAGE after switch 2 is ON \[{{V}_{f}}=\frac{{{C}_{1}}{{V}_{1}}}{{{C}_{1}}+{{C}_{2}}}=\frac{2V}{10}=0.2\,V\] Final energy in both the CAPACITORS \[{{U}_{f}}=\frac{1}{2}({{C}_{1}}+{{C}_{2}})V_{f}^{2}\,\,\,\,=\frac{1}{2}10{{\left( \frac{2V}{10} \right)}^{2}}=0.2{{V}^{2}}\] So energy dissipated \[=\frac{V{{}^{2}}-0.2{{V}^{2}}}{{{V}^{2}}}\times 100=80%\]