When the capacitor is charged by a battery of potential V, then energy stored in the capacitor,

${U_i} = {1 \over 2}C{V^2}$   ...(i)

When the battery is removed and another identical uncharged capacitor is connected in parallel

Common potential, $V' = {{CV} \over {C + C}} = {V \over 2}$

Then the energy stored in the capacitor,

${U_f} = {1 \over 2}\left( {2C} \right){\left( {{V \over 2}} \right)^2} = {1 \over 4}C{V^2}$   ...(ii)

$\therefore$ From eqns. (i) and (ii)

${U_f} = {{{U_i}} \over 2}$

that means the total electrostatic energy of resulting system will decreases by a factor of 2.