Derive terminal voltages for a BJT
Something does not make sense in the supplied answer, 43.3V on the collector implies Ic = (50V - 43.3V)/10k = 0.67mA, but Ie is 1mA due to the current source, so by Kirchoff Ib must be 0.33mA, making the voltage across Rb 6.6V (With the base negative with respect to ground, not impossible with a magic current sink in the emitter, but still).... 

Jonk has the right of it I think 33.3V Vc makes sense to me as well.

Here is how I would probably tackle this thing (But it has been a few years) 
:
Due to the 30V breakdown, Vb = Vc - 30V.

Then the collector current is the 1mA from the current source plus the current in the base pulldown resistor (Vc - 30V)/20k 
So Ic = 1mA + (Vc - 30V)/20k. From Kirchoffs laws concerning currents into and out of a junction.

But Vc can also be written in terms of the voltage drop across the collector load, Vc = 50V - 10k * Ic, so some substitution and trivial algebra should get you an answer.

Further due to the forward biased base-emitter diode we know that Ve = Vb - 0.7v.

That would be my take on it.