Post1 shows 68k base to emitter.
That value effectively sets the rail voltage that the amplifier will work at.
eg. Assume 1mA passes through the 220r emitter resistor. That gives a voltage drop of 220mVre.
Add on 600mVbe
That gives 820mV across the 68k.
and that leads to 5.66Volts across the 470k
The supply voltage equals 5.66+0.82V = 6.48Vdc when 1mA flows through the emitter resistor.
There is a small error in this simplified calculation. It ignores the base current, but you can compensate for that if you know the hFE of the transistor.
now repeat the calculation for a 5mA emitter current.
That will give a higher rail voltage. You could even try a 10mA emitter current, but that may make the transistor a bit too hot.
You can start with 27Vdc as your supply and work down to the emitter current.
Starting at that end will give a very high emitter current and almost certainly lead to overheating of the transistor.
Do that calculation and see what the predicted dissipation is.