Given: Length of the wire, \( L = 100 \, \text{cm} = 1.0 \, \text{m} \), Cross-sectional area, \( A = 2.0 \times 10^{-3} \, \text{cm}^2 = 2.0 \times 10^{-7} \, \text{m}^2 \) Resistance, \( R = 0.10 \, \Omega \) The formula for resistivity \( \rho \) is given by: \(\rho = R \cdot \frac{A}{L}\) Substituting the values: \(\rho = 0.10 \, \Omega \cdot \frac{2.0 \times 10^{-7} \, \text{m}^2}{1.0 \, \text{m}} = 2.0 \times 10^{-8} \, \Omega \cdot \text{m}\) Now, we can find the conductivity \( \sigma \): \(\sigma = \frac{1}{\rho} = \frac{1}{2.0 \times 10^{-8} \, \Omega \cdot \text{m}} = 5.0 \times 10^{7} \, \Omega^{-1} \cdot \text{m}^{-1}\) To convert to \( \text{cm}^{-1} \): \(\sigma = 5.0 \times 10^{7} \, \Omega^{-1} \cdot \text{m}^{-1} \times \left( \frac{1}{100 \, \text{cm}} \right) = 5.0 \times 10^{5} \, \Omega^{-1} \cdot \text{cm}^{-1}\) Thus, the electrical conductivity is: \(\sigma = 5.0 \times 10^{5} \, \Omega^{-1} \cdot \text{cm}^{-1}\).