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.
Soumee, I cannot identify any error in your calculation.
However, if you replace in the third line of your calculation the term (1+β) by β, your result will be identical to the expression as given in the book. As you know the current gain β is relatively large (mostly > 100), not a constant but dependent on Ic and - more important - equipped with large tolerances.
Therefore, we often simplify (1+β) to β (without expecting not acceptable errors) - and this seems to be the only reason for the discrepancy you have observed.
Best Answer
Unless you are good at iterative math, you'll have to make some simplifying assumptions:
From these, you can calculate the base current \$I_B=\frac{9\rm{V}-0.6\rm{V}}{220\rm{k\Omega}}\$ and collector current \$I_C=\frac{9\rm{V}-0.2\rm{V}-3\rm{V}}{330\Omega}\$, and then find \$\beta=\frac{I_C}{I_B}\$. If you find a transistor whose datasheet \$\beta_F\$ is larger than this value, then the circuit will work as advertised (once the transistor goes into saturation, its \$\beta\$ will drop to the calculated value). However, if you find a transistor whose datasheet \$\beta_F\$ is less, then it won't work as calculated -- and the collector current will be less, limited to \$I_C=\beta I_B\$.