Potential energy stored in the capacitor is U=1/2CV^2.
Dielectric with d=4 will increase the capacitance by 4.
Calculate the energy stored in the capacitor before and after
the dielectric is inserted - the difference will be the work
done to insert the dielectric. The justification why ΔU is the
work done in this case is somewhat fishy, at least from what I
can think of at the moment. There is another example, where you
have two connected capacitors with no battery and a dielectric
is inserted in one of them causing change of charge. Calculating
the total energy in both capacitors before and after the
insertion shows that there is a change in energy. In that case
it's easy to argue that the change of energy comes from work
because there are no other sources - the two capacitors are a
closed system. In your problem there is a transfer of energy
from the battery, so... What's the source of the problem?