Hii,

The solution this problem is as follows:

Charge on capacitor is

q = CV

when it is connected with another uncharged capacitor then

Vc = q1+q2/ C1+C2 = q+0/ C+C

Vc = V/2

intial energy

Ui = 1/2 CV^2

final energy

Uf = 1/2 C(v/2)^2 + 1/2 C(V/2)^2

= (CV^2/4)

Loss of energy = Ui - Uf

= CV^2/4

Therefore total electrostatic energy of resulting system  will decrease by a factor of 2.

Hope this helps.