Start by looking at the load resistor. If the BC243 is turned completely on, the emitter current would be limited to about 100 mA. In fact, looking at the path through the 10k resistor and the 2 base-emitter junctions, it's clear that the emitter of the BC243 will get to no more than about 13.6 volts (15 minus 2 times 0.7) and may be a little less, and this refines the maximum current estimate to about 13.5 volts / 151 ohms, or about 90 mA.
Is this reasonable? assuming a gain of about 100 (from the data sheet), the base current in the BD243 needs to be about 0.9 mA. Add 0.7 mA through the 1k resistor, and you need 1.6 mA from the BC337. Again, assuming a gain of 100, this means a base current of about 16 uA, which will require about 0.16 volts across the 10k resistor, which confirms a load voltage of about 13.5 volts.
Since the voltage across the BC337 is about half the voltage across the BD243, and the current is about 1/50, the BC337 dissipation will be about 1% and can be ignored. The VCE of the BD243 will be about 1.5 volts, and the current about 0.09 amps, so the dissipation will be about 135 mW.
With this said, the quiescent condition probably assumes that the 10k resistor is either open or connected to ground. In this case, the BC337 has no base drive, and the BD243 base is driven by leakage through the BC337. You'll need to look at the data sheet to determine this, then calculate the effects on the load and the transistor.