Single transistor Amplifier analysis (Q point values)
Task 1 - find Ib.

We know that VBE is 0.7V (given).{quick check - is this possible with the given values? - without the base connected R2 would produce a drop of about 0.75V  - this is a good result because we can always drop this voltage by taking current into the base}.

For VBE = 0.7 and R2 = 6.3k by Ohm's law I(R2) = 0.7/R2 = 0.11111 mA

The 11.3 volt drop (12 - 0.7) across R1 gives us a current of I(R1) = 11.3/93.7k = 0.120598mA

By Kirchoff's law the difference in current must be the base current 0.12 - 0.11 mA (2dp) = 0.01mA 

Task 2 - the collector current

We know that current gain is 100 (given)

So Ic = 100Ib  = 100 * 0.01mA = 1mA

Task 3 - find the voltage at the collector.

By Ohms law the voltage drop across Rc = Ic*Rc = 1(mA) * 6k = 6V

So the voltage at the collector (with respect to ground is 12 - 6 = 6V