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There are many answers to this problem, depending on your taste. Here's my answer:
With Q2 and Q3 running at 2mA, it would be convenient to run Q1 also at 2mA (so that its Vbe is similar). Thus, at "quiescent" conditions (i.e., VL=0V), RBIAS will have (15V-Vbe) across it, or roughly 14.3V. RBIAS=(14.3V/2mA)=7.2kOhm.
To ensure the voltage across Q1's Vce is 2Vbe, R1 should roughly equal R2 (neglecting finite beta), and most of the 2mA should flow through Q1. Let 200uA flow through R1 and R2. R1=R2=(0.7V/200uA)=3.5kOhm.
Next, run a simulation to confirm the accuracy of your calculations. My simulator shows 3mA flowing through Q2 and Q3, and 1.8mA through Q1. Not bad. Adjustments can be made to R1, R2, and RBIAS if you want to fine-tune things, but this is probably good enough.
The transistors are too weak to be used in that circuit. With an output of 10V p-p into the 10 ohm load then the peak output current is 500mA. But the maximum allowed output current of a 2N3904 and 2N3906 is only 200mA. They would also get very hot because the output power would be 1.25W and each transistor would dissipate about 0.5W.
If you use a lower output level, a higher load resistance or more powerful transistors then the datasheets of the transistors show the spec's that are used in the design of the circuit. Since transistors have a range of spec's then the idle current could be very low or very high unless an adjustment is used for each circuit that is made.
The typical current gain (hFE) at low current for a 2N3904 is 200. For a BD139 it is about 135 so of course the resistor values should be changed.
The BD139 has an old-fashioned case that is difficult to cool. Use a more modern transistor in a TO-220 case instead.
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