Off Grid Storage Capacitor Calculations
Plates that have an area of 0.6 sq m and a gap of 10 microns (0.01 mm) produce a capacitance of 1.59 uF when the dielectric constant is 3. With 20 such capacitors in parallel, that's a total capacitance of 31.8 uF. Your result seems to be off by a massive factor.
Web based parallel plate capacitor here: -


Is my math correct?

No - this was in error; 1x10^-11 for the gap. It should be 10E-6

running some led lights and some 500w loads for a few hours each
evening

Assume a kilowatt and assume the voltage is 240 volts. Assume that the voltage can be discharged to (say) 70% - that's a drop of 72 volts. Say you want the drop to occur over a period of 2 hours (7,200 seconds seems reasonable).
At the half way point (1 hour) the voltage will be about 200 volts and at a kW this is a current of 5 amps. The standard capacitor equation: -
$$I = C\cdot \dfrac{Δv}{Δt}$$
Rearranging for C: -
$$C = I\cdot\dfrac{Δt}{Δv} = 5\cdot\dfrac{\text{7200 seconds}}{\text{72 volts}}$$
$$\boxed{\text{This is a capacitance of 500 farads}}$$