To solve this problem, we will use the formula for the time period of oscillation of a magnetic needle in a magnetic field.$\\$The time period $T$ is given by:$\\ T = 2\pi \sqrt{\frac{I}{MB}} \\$Where:$\\$• $I = 9.8 \times 10^{-6}$ kg m$^2$ (moment of inertia)$\\$• $B = 0.049$ T (magnetic field)$\\$• $M$ is the magnetic moment of the needle$\\$Given that the needle performs 20 complete oscillations in 5 seconds, the time period $T$ is:$\\ T = \frac{5}{20} = 0.25$ s$\\$Substitute $T$ into the formula:$\\ 0.25 = 2\pi \sqrt{\frac{9.8 \times 10^{-6}}{M \times 0.049}} \\$Square both sides to eliminate the square root:$\\ (0.25)^2 = (2\pi)^2 \frac{9.8 \times 10^{-6}}{M \times 0.049} \\ 0.0625 = 4\pi^2 \frac{9.8 \times 10^{-6}}{M \times 0.049} \\$Rearrange to solve for $M: \\ M = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{0.0625 \times 0.049} \\$Calculate $M: \\ M = \frac{4 \times 9.8 \times 10^{-6} \times \pi^2}{0.0625 \times 0.049} \\ M = \frac{39.2 \times 10^{-6} \times \pi^2}{0.0030625} \\ M = 1280 \times 10^{-5}$ Am$^2 \\$Thus, the value of $x$ is $1280. \\$This corresponds to Option 2.