To solve this problem, we need to find the current density in the wire.$\\$Given:$\\$• Length of the wire $L = 10$ m$\\$• Radius of the wire $r = \frac{10}{\pi}$ mm$= \frac{10}{\pi} \times 10^{-3}$ m$\\$• Resistance of the wire $R = 10 \, \Omega \\$• Electric field strength $E = 10 \,$V/m$\\$First, calculate the cross-sectional area $A$ of the wire:$\\ A = \pi r^2 = \pi \left(\frac{10}{\pi} \times 10^{-3}\right)^2 \\ A = \pi \left(\frac{100}{\pi^2} \times 10^{-6}\right) \\ A = \frac{100}{\pi} \times 10^{-6} \,$m$^2 \\$Next, calculate the resistivity $\rho$ of the copper wire using the formula:$\\ R = \frac{\rho L}{A} \\ \rho = \frac{R \cdot A}{L} \\ \rho = \frac{10 \times \frac{100}{\pi} \times 10^{-6}}{10} \\ \rho = \frac{100}{\pi} \times 10^{-6} \, \Omega \cdot$m$\\$Now, use Ohm's law in terms of electric field and current density:$\\ J = \frac{E}{\rho} \\ J = \frac{10}{\frac{100}{\pi} \times 10^{-6}} \\ J = \frac{10 \pi \times 10^6}{100} \\ J = \pi \times 10^5 \,$A/m$^2 \\$Approximating $\pi \approx 3.14,$ we get:$\\ J \approx 3.14 \times 10^5 \,$A/m$^2 \\$Thus, the current density is approximately $10^5 \,$A/m$^2. \\$This corresponds to Option 4.