The maximum rated power of the LED is 2 mW and it is used in the...

Since the reverse-biased diode branch acts as an open circuit, the entire current through the series resistor $R_S$ flows through the LED:
$I_{\text{LED}} = I_{R_S} = 0.5\text{ mA}$

At the boundary condition where the LED dissipates its maximum rated power of $2\text{ mW}$:
$P_{\text{LED}} = V_{\text{LED}} \cdot I_{\text{LED}}$

$V_{\text{LED}} = \dfrac{P_{\text{LED}}}{I_{\text{LED}}} = \dfrac{2\text{ mW}}{0.5\text{ mA}} = 4\text{ V}$

Applying Kirchhoff's voltage law to the loop:
$V_{\text{input}} - V_{R_S} - V_{\text{LED}} = 0$

$5 - (I_{R_S} \cdot R_S) - 4 = 0$

$I_{R_S} \cdot R_S = 1\text{ V}$

$R_S = \dfrac{1\text{ V}}{0.5 \times 10^{-3}\text{ A}} = 2000\ \Omega = 2\text{ k}\Omega$

Hence, the correct option is $(2)\ 2$.