In an attempt to double the separation between the plates of a parallel plate capacitor while it still being connected to a battery, we observe that

For a parallel plate capacitor C = $\frac{{\in }_{0}A}{d}$ , Q = CV , $U_i=\frac{1}{2}C{V}^2$

When the separation is doubled , C'=$\frac{C}{2}, Q\text{'}=\frac{CV}{2}, U\text{'}=\frac{1}{4}C{V}^2$

In this process , $\left(CV - \frac{CV}{2}\right) i.e. \frac{CV}{2 }amount of charge flows through the battery from positive terminal to negative terminal .$

So work done by the battery , $W_b=- Charge flown \times emf = - \frac{C{V}^2}{2}$

Also if $W_e be the work done by the external agent , then W_b+W_e=(U\text{'}-U) + Heat generated .$

Since the circuit does not contain any resistor , heat generated is zero .