The BJT is in active region since V_{CE}
= V_{C} = 0.9V > 0.3V (therefore not in SAT), and a current is
flowing in the BJT (therefore not in cutoff).

I_{C} = 10µA therefore, the
base current is I_{B} = I_{C}/b = 10/60 = 0.167µA.

BJT in active: V_{BE} = V_{B
}= 0.7 (since V_{E} = 0 = ground)

Therefore, the voltage across R_{B} is
V_{C} – V_{B} = 0.9 – 0.7 = 0.2V.

Now, R_{B} = V_{RB}/I_{B}
= 0.2/0.167µ = 1.2 MOhms.

The voltage across R_{C} is V_{RC
}= 1.5 – V_{C} = 1.5 – 0.9 = 0.6V

Using KCL, the current flowing
through R_{C} is I_{RC} = I_{B} + I_{C} = 0.167
+ 10 = 10.167 µA.

This gives R_{B} = V_{RC}/I_{RC}
= 0.6/10.167µ = 3.6 MOhms.