When C1 is connected to voltage source, it is charged to a potential V and this will be stored as a potential energy in the capacitor given by U=21CV2 
When key is disconnected from battery and band c are connected, the charge will be transformed from the capacitor C1 to capacitor C2, then
The loss of energy due to redistribution of charge is given by ΔU=2(C1+C2)C1C2(V1−V2)2=2(C+C)C×C(V−0)2=41CV2[∵C1=C2]∴ Percentage loss =UΔU×100=21CV241CV2×100=50%