Mazy Loop

Applying KVL: $12I+2R_{3}=120$ $I=\frac{60-R_{3}}{6}$ And, $2R_{3}=20(I-2)$ $R_{3}=10(\frac{60-R_{3}}{6}-2)$

$\boxed{R_{3}=30}$

We know that Voltage R1 is same Voltage R2,R3 so The Current at R2 is 3 A. So the answer is 30 Ohm