but i am confused on how to extract the correct value.
The only way to find hFE for your BD139 is to measure it. It can be anywhere between the min and max given in the datasheet.
It is a good idea to design circuits so they work well with any value of hFE between min and max. This circuit is a good example. If Ic = 10mA, then Ib may be 10mA/40 = 0.25mA, or it may be 10mA/200 = 0.05mA.
As you calculated, current flow through R6 = 7.6mA, so current through the diodes may be 7.6mA - 0.25mA = 7.35mA, or it may be 7.6mA - 0.05mA = 7.55mA. The diodes don't care if the current is 7.35mA or 7.55mA. As long as the current through R6 is >> Ib, the circuit will work fine.
BTW, there is a problem with the load connected to the output of the current source. In theory Ic=Ie = Vbe/R5 = 10mA. However, that would mean the voltage drop across R7 = 10mA*2.7K=27V. That's impossible because the power supply is only 9V.
With the circuit as shown, hFE doesn't matter. The transistor is saturated, with Vce = about 0.05V.
Very rough calculation, assuming LED voltage = 1.7V:
Ie = 10mA
Ic = (9V - 1.7V - 0.7V) / 2.7K = 2.4mA
Ib = 10mA - 2.4mA = 7.6mA
current through diodes = 7.6mA - 7.6mA = about zero
If you made R7 higher than 2.7K, then Ie would be less than 10mA.