At the frequency of 5 MHz, the period of the oscillations is 1/5meg. That's a period of 1/5 microsecond.
There are 5 full cycles in one full microsecond, and there are 2.5 full cycles in a 0.5 us pulse.
You'll have to decide for yourself how damped a pulse of 2.5 cycles is, because the parameters of the definition are corrupted in the question.
Answer:
a) The charge of the capacitor is 4.25x10⁻¹¹C
b) The charge of the capacitor is 4.25x10⁻¹¹C because the battery is disconnected.
c) The potential difference across the plates is 18 V
d) The work is 7.64x10⁻¹⁰J
Explanation:
The capacitance of the capacitor is equal to:

A = 2 cm² = 0.0002 m²
d = 0.5 mm = 0.0005 m
Replacing:

a) The charge of the capacitor is equal to:
Q = C*V = 3.54 * 12 = 42.48 pC = 4.25x10⁻¹¹C
b) The charge is the same because the battery is disconnected (Q = 4.25x10⁻¹¹C)
c) If distance is increased, we have:

The potential is:

d) The work done is equal to:

F = 10 N
d = 20 m
θ = 0°
W = F (dot product) D = F * D * cos(angle between them)
W = FDcosθ
W = 10 * 20 * cos0 = 200 J