In the circuit below, the potential difference across the light emitting diode (LED) is 1.8 V when it is emitting light - AQA - A-Level Physics - Question 25 - 2017 - Paper 1

The voltage drop across the resistor can be calculated by subtracting the voltage across the LED from the total voltage:

$V_R = V_{total} - V_{LED} = 5V - 1.8V = 3.2V$

Using Ohm's Law, which states that $V = I imes R$, we can rearrange it to find the resistance:

$R = \frac{V_R}{I} = \frac{3.2V}{0.020A} = 160 \Omega$

The value of the resistor R is 160 Ω, therefore the answer is D.