Collector voltage for this transistor circuit
Well, at just first glance the answer is wrong - if \$I_0 = 600\,µA\$ then \$V_0 = I_0 * 10\,kΩ= 6\,V\$;   \$6\,V\ !=\ 12\,V\$ so the answers are contradicting.

Here is the solution:

Base current is what you calculated: \$ I_b = 2.5 * 10^{-6}\,A\$

Collector current: \$ I_b*\beta = 2 * 10^{-4}\,A\$

Now \$ V_0 \$ is simply: \$ 20\,V - i10^4\,Ω \$ (\$i\$ is the current of the top most resistor)

Currents equation: \$i = i_c + i_0 \$   (\$ i_c\$ is collector current)

Next we put current equation into \$V_0\$ equation:
\$V_0 = 20\,V - (i_c + i_0)10^4\,Ω\$

\$I_0\$ is \$ \frac{V_0}{10^4}\$

So we combine above three equations into: \$V_0 = 20\,V - (i_c + \frac{V_0}{10^4\,Ω})10^4\,Ω \$

\$2V_0 = 20\,V - i_c*10^4\,Ω\$

\$V_0 = 9\,V \$

and finally

\$I_0 = \frac{9\,V}{10^4\,Ω} = 900\,µA \$