The diameter of the disks is 1.32 cm.
If n electrons each of charge e are transferred from one disc to another, calculate the total charge Q transferred from one disc to the other using the expression,

Substitute 1.50×10⁹ for n and 1.6×10⁻¹⁹C for e.

The potential difference V between the disks separated by a distance d is given by,

here, E is the electric field.
Substitute 2.00×10⁵N/C for E and 0.50×10⁻³m for d.

The capacitance C of the capacitor is given by,

The capacitance of a parallel plate capacitor is given by,

Here, ε₀ is the permittivity of free space and A is the area of the disks.
Rewrite the expression for A.

the area A of the disks is given by,

Here, D is the diameter of the disk.

The diameter of each disc is found to be 1.32 cm.