Biasing a crappy FM transmitter circuit

An answer:

\begin{array}{rcl} V_{CE} & = V_{CC} - V_{R_{C}} \\ = V_{CC} - I_{C} \cdot R_{C} \\ I_{C} = βfe I_{B} \\ I_{B} = \frac{V_{CC} - V_{B}}{R_{B 1}} \\ V_{CE} & = V_{CC} - βfe \frac{V_{CC} - V_{B}}{R_{B 1}} R_{C} \end{array}

yields:

\begin{array}{rcl} I_{C} & = 0.84 mA \dots 6.7 mA \\ R_{C} & = 1 k Ω \end{array}

and hence gives

\begin{array}{rcl} V_{R_{C}} & = 0.84 \dots 6.7 V \\ V_{CE} & = 8.16 \dots 2.3 V \end{array}

which is fine for biasing purposes.