At first I introduced a new current Iq which flows through R2. Having
that done I know that Iq=UBE/R2.
This is incorrect; the voltage across R2 is \$U_{BE} + I_E R_E\$
Also, I suspect that the values of R1 and R2 should be in \$k\Omega\$ and the value of \$R_E\$ is suspiciously high.
Regardless, there's a step by step approach to finding \$I_B\$.
Form the Thevenin equivalent circuit looking out of the base:
\$U_{BB} = U_{CC} \dfrac{R_2}{R_1 + R_2}\$
\$R_{BB} = R_1 || R_2\$
Now, write the KVL equation around the base-emitter loop:
\$U_{BB} = I_B R_{BB} + U_{BE} + I_E R_E\$
Using the relationship:
\$I_E = (\beta + 1) I_B\$
Substitute and solve:
\$I_B = \dfrac{U_{BB} - U_{BE}}{R_{BB} + (\beta + 1)R_E}\$
You can ignore this if you like, but you ought to, before turning in or publishing an answer, do a sanity check to make sure that, on the face of it, your answer isn't hopelessly, impossibly wrong.
For example, consider the answer you give for the base current and the implication of it. If the base current were 0.2A, as you've calculated, the emitter current, which is 501 times the base current, would be an enormous 102A.
It's always good to do a sanity check on your answer. Even if \$U_{CE}\$ were zero, the emitter current could not be any larger than:
\$I_{E_{max}} = \dfrac{U_{CC}}{R_C + R_E} = 984\mu A\$
This places an upper bound on the base current which is:
\$I_{B_{max}}= \dfrac{I_{E_{max}}}{\beta + 1} = 1.96\mu A\$
So, by making a very quick calculation, you have a good sanity check for any answer you may come up with.
Pretty close.
You should give the PNP a lot more base current to make sure it saturates. Use a forced beta of 10-20, which means that the resistor R2 should be more like 470R.
The value of R3 is not critical, 10K would be fine, so would 20K.
You don't need a Darlington to drive the PNP transistor, a regular NPN will be fine. Use a base resistor of more like 10K. The base resistor limits the current- it is determined by the desired base current, not by the gain of the transistor- you do always need it.
Keep in mind these Darlington things won't switch instantly and if you don't allow some dead time in the drive firmware you'll get 'ghosting'.
Best Answer
Based on the numbers you have, and assuming you want the collector to rest at 50% of the supply voltage. then Ohm's Law:
E = I x R
R = E / I
R2 = (Supply/2) / 0.005 A
R2 = 5 / .005
R2 = 1 K
Note that the circuit will not actually work. Biasing the base with a single resistor is called dangle biasing, and is severely unstable. The gain of the transistor changes with temperature, with changes in collector current, and from part to part.