Correct Option: B

Since transistor in the saturation. op-amp having, R_f = 0

Vo = 2		1 +	Rf		= 2 V
			R1

Negative voltage drop between base and collector terminal due to virtual ground,
V_A = 2 V

so, current in 2 kΩ resistor =	12 – 2	= 5 mA = I
	2 kΩ

Now, since voltage at C is 2 V
so, 2 = I × R_L
2 = 5 × 10^–3. R_L
or R_L = 400 Ω