Strictly speaking, the circuit has 5 nodes, at points labelled a, b, c, d and f. If you use the modified nodal analysis to solve the circuit, you'll apply KCL at all nodes but one (usualy the reference one) to end up with a system of lineal independent equations. So, you'll apply KCL at nodes a, b, c and d to determine the (initially) unknown voltages at these nodes.
However, this is modified by the presence of voltage sources connected to ground at nodes a and c. Due to the presence of these voltage sources, the voltages at nodes a and c are not unknowns, so you have only 2 "real" unknowns (voltages at b and d), and you have to write only 2 KCL. If you have to choose, you would like to write KCL at b and d, because writing the KCL at a and c involves the currents flowing through the voltage sources, and you don't know how to express these currents in terms of the node voltages, so you avoid writing KCL equations at nodes having voltage sources connected to ground.
So finally you have to write the KCL at b and d to solve the circuit, and you can also get rid of the KCL at d if you add the impedances of the resistor and the capacitor together to have 10-10j Ohm.
So your first equation (KCL) is ok, and all you have to do from that is to express the currents in terms of node voltages and impedances:
$$I_1+I_2−I_3−I_4=0$$
$$I_1 = \frac{E_1-V_b}{-10j}$$
$$I_2 = \frac{E_2-V_b}{+20j}$$
$$I_3 = \frac{V_b}{10-10j}$$
$$I_4 = \frac{V_b}{10}$$
And if you substitute these currents in the first KCL and solve for Vb you obtain:
$$V_b = \frac{1-j}{1-3j} * (2E_1-E_2) = \frac{1}{2-j} * (2E_1-E_2)$$
Knowing Vb (and E1 and E2) allows you to easily determine all other circuit variable.
The answer given is clearly wrong.
20 volts across a 10K resistor will give 2 mA, so \$I_0\$ can't possibly be 600 mA.
If we omit the transistor, the two 10K resistors form a voltage divider, dividing the 20 volt supply in half, making \$V_0\$ 10 volts.
The transistor will draw some current in parallel with the lower 10K, so \$V_0\$ must be less than 10 volts.
Best Answer
One current supply cannot circulate current in the other mesh. There won't be any steady state electrical charge transfer between the meshes. Each meshes are independent circuits. The connection at point-A only keeps the meshes at the same potential; just like putting a common ground between two circuits.
There are two independent mesh current equations, each of them having only one unknown variable. Therefore, I1=1A and I2=5A.