Correct choice is (c) (10∠0^o)/3
For explanation: The general equations are YaaVa+YabVb = I1, YbaVa+YbbVb = I2. We get Yaa=1/3+1/j4+1/(-j6) and the self admittance at node a is the sum of admittances connected to node a. Yab=-(1/(-j6)). I1 = (10∠0^o)/3=x.