Biasing in a metal detector circuit

An answer:

\begin{array}{rcl} I_{C} & = \frac{βfe}{βfe + 1} \frac{V_{E}}{R_{4}} \\ V_{E} = V_{B} - V_{BE} \\ V_{B} = V_{CC} \frac{R_{2}}{R_{1} + R_{2}} - (R_{!} ∕ ∕ R_{2}) \cdot I_{B} \\ I_{B} = \frac{I_{C}}{βfe} \\ ⟹ \\ I_{C} & = \frac{βfe}{βfe + 1} \frac{V_{CC} \frac{R_{2}}{R_{1} + R_{2}} - R_{!} ∕ ∕ R_{2} \frac{I_{C}}{βfe} - V_{BE}}{R_{4}} \\ = \frac{βfe}{βfe + 1} \frac{V_{CC} \frac{R_{2}}{R_{1} + R_{2}} - V_{BE}}{R_{4} + \frac{1}{βfe + 1} \cdot R_{!} ∕ ∕ R_{2}} \end{array}

Substituting the numerical values yields $I_{C} = 0$ because $V_{B} \approx 0$ . Most likely $R 2$ should be 330k $Ω$ .