Solving a transistor circuit and determine the base current
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.