If we apply the KVL in the big loop (left and middle loop), we get:
Vs = 10V = - 20*i2 – 30*i1 which is 1 = - 2*i2 – 3*i1
And with the current source, we get
Is = 0.5 A = i2 – i1
We have then two equations with 2 unknowns, which give:
I1 = - 3/5 A and i2 = - 1/10 A
The voltage is thus v2 = 20*i2 = - 2V