How much energy has been converted by the cell in the charging process?

Well, emf is "energy per coulomb", so if you multiply emf by Q ...

Well, emf is "energy per coulomb", so if you multiply emf by Q ...

This gives me the correct answer. $6 \times (6 \times 10^{-4}) = 3.6 \times 10^{-3}J$

Yes, in fact that's right. The question asks about the energy converted by the cell, not the energy stored in the capacitor - I misread the question. Someone really ought to downvote me to take away that evil, undeserved "+" that someone clicked.

Surprisingly, the uncharged capacitor does not resist the current arriving at one plate or leaving the other.

The whole point of this exercise is that the cell converts QV joules of energy in the process, as I said in my original reply and was asked by the original question, and the capacitor, when charged, only has 0.5QV joules of energy stored in it.

So where did the missing energy go?

Classic physics problem.

If we assume that there is series resistance $R$ between the capacitor and cell, then the cell transfers $Q$ coulombs through it, at an average voltage of $\frac{V}{2}$, (since the voltage drop across $R$ starts at $V$ and falls linearly to 0), giving an energy dissipation of , which is precisely half of the energy, the other half being stored in the E-field of the capacitor.

If we assume that there is no series resistance between the capacitor and cell, then this is too unphysical to analyse properly (since we'd be assuming a perfect voltage source, infinite current to bring the capacitor up to full charge instantly => infinite radiative losses, etc).

Have I solved the classic physics problem?

Yes of course. It is lost in the resistance of the connection between the two. It doesn't matter what the value of the resistance is, either. But there must always be some resistance, somewhere, in the real world.
So let's say you connect the cell directly to the plates using no wires at all.
What happens now?
There is the internal resistance of the cell. The charge would have to flow in and along the capacitor plates, and these must have some resistance.
So yes. It's a classic little A Level teaser.

Surprisingly enough that's the first time I've seen it presented as a teaser, but it's a nice A level standard question.

Are you familiar with the much trickier two-capacitor conundrum?