In the circuit shown below the current in the 1Ω resistor is:

0.13 A from Q to P

The correct answer is Option (3) → 0.13 A from Q to P

Using Kirchoss voltage across ABQP,

$-6+3I_2+1(I_2-I_1)=0$

$⇒4I_2-I_1=6$   ...(1)

Using Kirchoss voltage across QDCP,

$-6+2I_1-(I_2-I_1)+3I_1=0$

$⇒-I_2+6I_1=9$   ...(2)

Using (1) and (2)

$I_2=0.13A$

Direction Q to P, Since $I_2>I_1$