The RC time constant is 1 second.
Thank you for the schematic diagram. The math is like this after the switch is released:
There are three nodes : S1 Anode GND
The LED Anode has about 2v as a diode drop. Treat it as a voltage source of 2v.
The resistor has 3 volts time = t = 0 seconds
The peak current at t=0 is i(0) = v/r = 3/10k = 0.3mA
The current decays exponentially
i(t) = i(0) * e^(-t/rc)
rc = 10k * 100uF = 10^4 * 10^-4 = 1 second
e = 2.7182 a constant from science education
e^-1 = 1/e = .632
e^-2 = 1/(e*e) = .632 * .632 = .40
e^-4 = .16
e^-8 = .0256
after 1 second i = .3mA * .632 = .189 mA
after 2 seconds i = .3mA*.16 = 48 micro amps (uA)
after 8 seconds i = .3mA*.0256 = 8uA
Conclusion : your math was wrong. The LED will be dim with .3mA.
Its power is 2v*.3mA = 0.6 milliwatts
This is a dim bulb.