First, let's consider the capacitor with the dielectric between its plates. The charge on the capacitor is Q, its capacitance is C (which includes the effect of the dielectric), and the potential difference (voltage) across its plates is V. According to the formula for the energy stored in a capacitor :

$$U = \frac{1}{2} CV^2$$

we find that the energy is equal to half of the product of the capacitance and the square of the voltage.

Now, consider the process where the dielectric slab is removed from the capacitor. When the dielectric is removed, the capacitance of the capacitor decreases. However, because the capacitor is not connected to anything that can supply or absorb charge, the charge Q on the capacitor stays the same. Since the charge stays the same but the capacitance decreases, the voltage V across the capacitor must increase to keep Q = CV true.

The energy of the capacitor without the dielectric is still given by :

$$U = \frac{1}{2} CV^2$$

but now C is smaller and V is larger. However, because both C and V² change in such a way that their product remains constant, the energy of the capacitor doesn't change when the dielectric is removed.

Therefore, the energy of the capacitor before the dielectric is removed is the same as the energy of the capacitor after the dielectric is removed. When the dielectric is reinserted, the process is just reversed, so again no net work is done.

So the total work done by the system in the process of removing the dielectric and then reinserting it is zero. This is because work is the transfer of energy, and in this case, the energy of the system (the charged capacitor) does not change. Therefore, no energy is transferred, so no work is done.