First, there is a mistake in your initial equations. Ic = Beta * Ib and therefore Ib + Ic = (1 + Beta)Ib, NOT (1 + Beta)Ic - which will make for a huge difference quantitatively !
Furthermore, I would like to answer this with some added commentary that applies to real world design. First, the voltage at the base must be the rated zener voltage (12V) so long as the zener diode has enough ballast current through it. This is the purpose of the 220 Ohm resistor betwenn Vi and the zener/base node. Without it, there would be no current in the zener because for the transistor to be in forward-active mode (which it needs to be in for this circuit to work) current must flow INTO the base, NOT out of it. And therefore the 220 Ohm resistor ALSO acts to bias the transistor (i.e., provides current INTO the base). The bulk of the current in the 220 Ohm resistor - but not quite all of it - is biasing the zener diode. The remainder is the base current. Using KVL & KCL, the current in the 220 Ohm Resistor (call it Rb; 'b' for 'bias' or 'ballast'):
I[Rb] = (Vi - Vz)/Rb = 8V/0.22Kohms = 36.36mA
The current in Rb that diverts into the base is:
Ib = Ic/Beta = Ic/50 = (Ie - Ib)/50; therefore,
Ib = Ie/51.
Now, Ie is just V0 / RLoad = 11.3V / 1.0Kohms = 11.3mA. . . . because . . . . .
When a transistor is in forward-active mode, assuming no component power ratings are being exceeded, there is a FORWARD diode drop from Vb to Ve, i.e.,
Ve = Vb - 0.7V. . . . . therefore . . . .
Vo = Ve = Vb - 0.7V = Vz - 0.7V = 12V - 0.7V = 11.3V and therefore:
Ie = ILoad = IRLoad = 11.3V / 1Kohms = 11.3mA.
. . . . . and now, by the way, we can calculate Ib:
Ib = Ie/51 = 11.3mA/51 = 222uA which as expected is a tiny fraction of the ballast current, or the current in the zener diode. Finally . . . . .
Iz = I[Rb] - Ib = 36.363mA - 0.222mA = 36.142mA
A couple of interesting points, again practical design matters.
First, note that the power we are asking Rb to dissipate is:
P[Rb] = Rb * V[Rb]^2 = 220 ohms * 64 V*V = 0.29W . . . .
which is more than a quarter watt. So if you are going to build this you'd better
choose at least a half-watt resistor for Rb. Power dissipated by the transistor is approx
Px = (Vc-Ve) * Ie = (20-11.3)V * 0.012 A = 0.1W = a tenth of a Watt.
Does something seem weird here ? There is more current in the ballast resistor than there is in the load ! And more power dissipated in the ballast resistor than in the transistor. I am not saying this is wrong, it will work, but as a practical matter, it is wasteful; it would be better to use a zener that operates well with only about one-tenth of the current in this circuit, or about 3.6mA. Such things do exist. That way, you can use a much lower value AND power Rb than 1/2Watt ("an exercise for the student" - what would be the new Rb value ?) and it will work fine because Ib is still a small fraction of Iz (about 0.064 or one-sixteenth - please confirm) so the transistor will still regulate the output voltage just fine. Rload = 1K is not much of a load in this circuit for discreet components. In fact you can keep this new lower Rb value, and lower RLoad substantially, thereby INcreasing ILoad, and so long as you do not exceed the power rating of the transistor, that would be fine. ALSO, ILoad is limited by Beta, because with excessive ILoad, Ib requirement will become too large, robbing too much current from the zener, and the zener will no longer regulate. Those two differences brings us closer to a real world practical example. For a circuit presented in a textbook to illustrate the principals of analysis, it's OK, but from a practical viewpoint, it should have the adjustment I mentioned.
One more important point. Note that Vi = 20V "unregulated" means that Vi varies somewhat. In the real world this could be quite a bit, in fact. As Vi varies so do a lot of other things like many of the currents we calculated earlier. The trick is that the zener must have a very flat shelf in its V/I transfer curve, which means Vz = rock solid 12V no matter what Iz might be. It will never be perfectly flat, so Vz and therefore Vo will vary somewhat with varying Vi. But the variance will be much attenuated. If you look at specs for power supplies, you will see this attenuation factor. The more attenuation, the better.
One more question. Why not just use a zener without a transistor (assuming you want Vo = 11.3V and you can find a zener at that rating) ? In fact, there are some real-world applications that do just that, when the load requirement is not much. You probably already saw that in a previous lecture/book chapter.
Ans: For load current requirements in the neighborhood of the zener ballast current or higher, the answer is because the transistor, with its remarkable current amplification property - it's all about the Beta - isolates the load from the zener. Do you see that ?