Right option is (b) 2.5 A
Easiest explanation: Applying KCL, we get, I1 + 5 = I2 + I3
∴ \(\frac{10 – V_1}{1} + 5 = \frac{V_1}{2} + \frac{V_2}{6}\)
∴ 30 – 3V1 + 15 = 3V1 + V2
∴ 6 V1 + V2 = 45
From voltage source, V2 – V1 = 10
Now, 7 V1 = 35, V1 = 5 V
And V2 = 15 V
∴ IX = \(\frac{V_2}{6} = \frac{15}{6}\) = 2.5 A.