In the case of self-inductance of the solenoid, magnetic flux (ϕ) is directly proportional to the current (i) flowing in the coil. The flux through each turn is equal to the flux linkage.
The relation is given as Nϕ=Li, here N is the number of turns and L is inductance.
So putting the given values, we get
1000×4×10−3=L×4
Thus, the self inductance is 1H.