Answer
Germanium.
Work Step by Step
Use the resistor color code (Figure 18-12) to find that R = $24\times10^1=240\Omega$.
Equation 18–3 also gives the resistance. The cross-sectional area $A=\pi r^2=\frac{\pi d^2}{4}$.
$$R=\rho\frac{\mathcal{l}}{A}=\rho\frac{4 \mathcal{l}}{\pi d^2}$$
$$240\Omega=\rho\frac{4 (9.00\times10^{-3}m)}{\pi (2.15\times10^{-3}m)^2}$$
$$\rho=97\times10^{-3}\Omega \cdot m$$
Of the materials listed in Table 18–1, germanium has a resistivity in this range and is the likeliest candidate.
