Note that we can use I_B for the mesh current on the left since it is the only current on the left side of the circuit.

I_B is just itself.
V_CE = 40 K I_C

Define the current source in terms of the mesh currents.

200 I_B = - I_C (Eq. 1)

This is one of our final mesh equations.
Removing the current source we see that there is no mesh on the right, thus only one KVL equation on the left.

1K I_B + 4K I_B + 10_-5 (40 K I_C) - 10 m = 0

Inserting the equivalent of I_C from Eq. 1 gives

5K I_B + 400m (- 200 I_B) = 10m
4920 I_B = 10 m
I_B = 2.0325 mA

Eq. 1 now gives

I_C = - 200 I_B = -200 (2.0325m)
I_C = - 406.5 mA

Finally, since V_CE = V_OC,

V_OC = 40K I_C
VOC = -16.26 V

Using the Thevenin equaivalent resistance found in problem 5, the Thevenin equivalent circuit is thus

OR

To check this result, we can find I_SC and calculate R_Th using V_OC.

First note that V_CE is zero, since it is shorted out, thus the voltage across the dependent voltage source is also zero.
KVL around the left side then yields

5K I_B = 10 m
I_B = 2 mA

Now, since the current through the 40 K resistor must be zero (why?), KCL immediately implies that

I_SC = - 200 I_B
I_SC = - 400 mA

Thus

R_Th = V_OC / I_SC = -16.26 / -400m
RTh = 40.65 KW

which agrees with our earlier result.