To solve this problem, we will use the formula for the magnetic potential energy stored in an inductor.$\\$The formula is given by:$\\ U = \frac{1}{2} L I^2 \\$Where:$\\$• $U$ is the magnetic potential energy stored in the inductor (in joules),$\\$• $L$ is the inductance of the inductor (in henries),$\\$• $I$ is the current through the inductor (in amperes).$\\$Given:$\\ U = 25$ mJ$= 25 \times 10^{-3}$ J$\\ I = 60$ mA$= 60 \times 10^{-3}$ A$\\$Substitute the given values into the formula:$\\ 25 \times 10^{-3} = \frac{1}{2} L (60 \times 10^{-3})^2 \\$Simplify the equation:$\\ 25 \times 10^{-3} = \frac{1}{2} L \times 3600 \times 10^{-6} \\ 25 \times 10^{-3} = \frac{1}{2} L \times 3.6 \times 10^{-3} \\$Multiply both sides by 2 to solve for $L: \\ 50 \times 10^{-3} = L \times 3.6 \times 10^{-3} \\$Divide both sides by $3.6 \times 10^{-3}: \\ L = \frac{50 \times 10^{-3}}{3.6 \times 10^{-3}} \\ L = \frac{50}{3.6} \\ L \approx 13.89$ H$\\$Therefore, the inductance of the inductor is approximately $13.89$ H.$\\$This corresponds to Option 4.