Answer:
For a resistance (R) of 1000 MΩ and capacitance (C) of 2.179x10⁻³μF
Step-by-step explanation:
The charge on the discharging capacitor is equal:

If t = 1 s, then:

Matching both equations:

If t = 1

The capacitance for a resistor of resistance of 1000 MΩ is equal:
