1. Interpret the diode model

   Each ideal diode has:

   • zero resistance in reverse bias branch (open circuit),
   • forward resistance $= 50\,\Omega$ when conducting.

2. Identify which diodes conduct

   From the given circuit, only one diode becomes forward biased in the conducting path with the $100\,\Omega$ resistor, while the other diode is reverse biased and does not conduct.

   So the effective series resistance in the active loop is: $R_{\text{eq}} = 100 + 50 + 150 = 300\,\Omega$

   where:

   • $100\,\Omega$ is the given resistor,
   • $50\,\Omega$ is the forward resistance of the conducting diode,
   • $150\,\Omega$ is the other series resistor in the loop.

3. Apply Ohm's law

   Battery voltage is $6\,\text{V}$, hence the circuit current is: $I = \frac{V}{R_{\text{eq}}} = \frac{6}{300} = 0.020\,\text{A}$

4. Current through the $100\,\Omega$ resistor

   Since the resistor lies in the same series conducting loop, the current through it is the same: $I_{100\Omega} = 0.020\,\text{A}$

5. Option matching

   $0.020\,\text{A}$ corresponds to Option D.

Comparison with stored answer: Stored correct answer is D, which matches the derived result.