To solve this problem, we need to find the self-inductance of the solenoid.$\\$Given:$\\$• Number of turns $N = 1000 \\$• Current $I = 4 \,$A$\\$• Magnetic flux linked with each turn $\Phi = 4 \times 10^{-3} \,$Wb$\\$The formula for the self-inductance $L$ of a solenoid is given by:$\\ L = \frac{N \Phi}{I} \\$Substitute the given values into the formula:$\\ L = \frac{1000 \times 4 \times 10^{-3}}{4} \\$Simplify the expression:$\\ L = \frac{4000 \times 10^{-3}}{4} \\ L = \frac{4}{4} \\ L = 1 \,$H$\\$Therefore, the self-inductance of the solenoid is $1 \,$H$. \\$This corresponds to Option 2.