As the voltage between the capacitor's plates decreases, so should the
current flowing through the circuit.
I don't follow your reasoning here. Recall that, for an ideal capacitor, we have:
$$i_C = C\frac{dv_C}{dt}$$
In words, the current through the capacitor is proportional to the rate of change of the voltage across, not the instantaneous value of the voltage.
So, for example, if the voltage across the capacitor is sinusoidal
$$v_C = V \sin\omega t$$
the current is
$$i_C = \omega CV \cos \omega t$$
which means (1) that the maximum current (magnitude) occurs when the voltage is zero and (2) that the maximum voltage (magnitude) occurs when the current is zero.
Now, for this simple LC circuit, the voltage across the capacitor is identical to the voltage across the inductor:
$$v_C = v_L$$
thus,
$$i_C = C\frac{dv_L}{dt}$$
For an ideal inductor, we have:
$$v_L = L\frac{di_L}{dt}$$
But, the inductor current is
$$i_L = - i_C$$
thus,
$$i_C = -LC\dfrac{d^2i_C}{dt^2}$$
which means that the current is sinusoidal
$$i_C = A \sin \omega t + B \cos \omega t $$
where
$$\omega = \frac{1}{\sqrt{LC}}$$
Since, in your example, the initial current is zero and the initial voltage is $V$, we have
$$i_C(t) = -\frac{V}{\omega L} \sin \omega t$$
In real life an inductor consists of a coil of wire (with or without a laminated iron core). So a real inductor has both resistance and inductance. If you double the inductance by increasing the length of wire on the coil, then the resistance will increase (roughly 1.4 times). All the same, your question should have made it clear that you were dealing with a real inductor, not an ideal or 'pure' inductor. And it should also have told you that the inductance is doubled by adding more turns to the coil.
Your question should also have made it clear whether it wanted a comment about electron drift velocity just after the inductor is connected to the battery, or after a steady current is achieved. Your answer is suited to the former (though it looks as if you should include a resistive term in your equation); their answer will also apply in the latter case, though, as I've said, I think they should have warned you that it was not a pure inductor.
You'll have gathered that I don't think much of the question!
Best Answer
When the switch is opened, the inductor energy is dissipated over time via the internal inductor resistance and the external parallel resistance.
The stored energy is radiated away as heat, heating both the inductor and the external parallel resistor.
If the external resistance is much larger than the internal inductor resistance, the majority of the stored energy will be dissipated by the external resistor (in the limit of zero inductor resistance, all the energy is dissipated by the external resistor).
Conversely, if the internal resistance is much larger, the majority of the energy will be dissipated by the inductor (in the limit of zero external resistance, all the energy is dissipated by the internal inductor resistance).