Let ib = 6 mA
We have Vb = 100*(ib – ia) = 100*(6. 10-3 – ia)
On the right side we have the KVL and:
Vb = 3*Vb + 250*ia
Thus - 2*Vb = 250*ia so vb = -125*ia
So with the first equation we get:
-125*ia = 100*(6. 10-3 – ia) = 0.6 –100*ia
thus - 25*ia = 0.6 so ia = - 0.024 = - 24 mA