Correct Option: D

From the given information, we can make the equivalent circuit as shown below:
From the above circuit KVL in loop 1
20–0.7 –0.3–5.6 kΩ × I_D1 = 0

ID1 =	19 V	= 3.393 mA
	5.6 kΩ

and I3kΩ =	0.3 V	= 0.09 mA
	3.3 kΩ

So, I_D2 = I_D1 – I_3kΩ = 3.303 mA