Since the reverse-biased diode branch acts as an open circuit, the entire current through the series resistor RS flows through the LED:
ILED=IRS=0.5 mA
At the boundary condition where the LED dissipates its maximum rated power of 2 mW:
PLED=VLED⋅ILED
VLED=ILEDPLED=0.5 mA2 mW=4 V
Applying Kirchhoff's voltage law to the loop:
Vinput−VRS−VLED=0
5−(IRS⋅RS)−4=0
IRS⋅RS=1 V
RS=0.5×10−3 A1 V=2000 Ω=2 kΩ
Hence, the correct option is (2) 2.