Resistive circuit analysis with dependent current and voltage source
The first thing you should notice is the \$R_1\$ resistor is connected directly in parallel with \$V_{S2}\$ voltage source. 

All this means that we already know the \$I_A\$ current value. 

$$I_A = \frac{V_{S2}}{R_1} = \frac{12V}{6k\Omega} = 2mA $$

And this current will have no effect on the remaining part of the circuit because the voltage at node \$B\$ is fixed and it is equal to \$12V\$. 

Therefore your circuit will now look like this: 


Loop B KVL 

$$V_{S1} - I_B R_2 - (I_B + I_C)R_3 - V_{S2} - (I_B + I_C)R_6 = 0$$

The KVL equation for loop C is not needed because we have a current source in this 

loop hence \$I_C = I_1 = 1mA\$ 

Solving this will give us the answer: 

$$I_B = I_{R2} = -4.4mA$$ 

And this minus sign tells us the \$I_B\$ current is flowing in the opposite direction to the direction I have marked on the diagram. 

We can do the nodal analysis also.

We know that the voltage at node B is equal to 12V. Hence the nodal equation for node A can look like this:

$$ \frac{V_A - V_{S1}}{R_2} + \frac{V_A - V_{S2}}{R_3 + R_6} - I_1 = 0$$

And the solution is $$V_A = 20.8V$$

And we are done.