A parallel-plate capacitor having plate area A and plate separation...

Initial capacitance $C_i = \dfrac{\varepsilon_0 A}{d}$

Initial charge on the capacitor $Q = C_i V$

Initial energy of the system $U_i = \dfrac{1}{2} C_i V^2 = \dfrac{\varepsilon_0 A V^2}{2d}$

Since the battery is disconnected, the charge $Q$ remains constant.

When the dielectric slab is inserted, the final capacitance becomes $C_f = K C_i = \dfrac{K \varepsilon_0 A}{d}$

Final energy of the system $U_f = \dfrac{Q^2}{2 C_f} = \dfrac{(C_i V)^2}{2 K C_i} = \dfrac{C_i V^2}{2K} = \dfrac{\varepsilon_0 A V^2}{2Kd}$

The work done on the system by the external agent is equal to the change in the potential energy of the system.

$W = \Delta U = U_f - U_i$

$W = \dfrac{\varepsilon_0 A V^2}{2Kd} - \dfrac{\varepsilon_0 A V^2}{2d}$

W = \dfrac{\varepsilon_0 A V^2}{2d} \left(\dfrac{1}{K} -1 \right)