Verified answer

When the cap is switched, to simplify the calculation, you can subtract 6 volts from the cap voltage and 6 volts from the battery, then you have an uncharged battery, a 3M resistor, and a 6 volt battery, charging to 3 volts.

voltage on a cap, charging

v = v₀[1–e^(–t/τ)]

v₀ is the battery voltage

v is the voltage after time t

R is resistance in ohms,

C is capacitance in farads

t is time in seconds

RC = τ = time constant

τ = 4µ x 3M = 12 sec

v = v₀[1–e^(–t/τ)]

3 = 6[1–e^(–t/12)]

1–e^(–t/12) = 0.5

e^(–t/12) = 0.5

–t/12 = –0.693

t = 8.32 sec