Answer:
The capacitance of your capacitor is 5.476 x 10⁻⁵ μF
Step-by-step explanation:
Given;
diameter of the aluminum pie plates = 16 cm = 0.16 m
separation distance, d = 3.25 mm = 0.00325 m
voltage across the parallel plates = 6 V

where;
C is the capacitance of your capacitor
ε is the permittivity of free space = 8.85 x 10⁻¹² (F/m)
d is separation distance
A is the area of the plate = ¹/₄ (πd²) = 0.25 (π x 0.16²) = 0.02011 m²

Therefore, the capacitance of your capacitor is 5.476 x 10⁻⁵ μF