In the case of self-inductance of the solenoid, magnetic flux $\left(\varphi \right)$ is directly proportional to the current $\left(i\right)$ flowing in the coil. The flux through each turn is equal to the flux linkage.

The relation is given as $N\varphi =Li$, here $N$ is the number of turns and $L$ is inductance.

So putting the given values, we get

$1000\times 4\times {10}^{-3}=L\times 4$

Thus, the self inductance is $1 \mathrm{H}$.