The total number of turns and cross-section area in a solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to:

For an air-core solenoid the self-inductance is given by the well-known relation

$$L = \mu_0\,n^2\,A\,l,$$

where  $$\mu_0$$  is the permeability of free space,  $$n$$  is the number of turns per unit length,  $$A$$  is the cross-sectional area of the solenoid, and  $$l$$  (here written as $$L$$ in the question) is the actual length of the solenoid.

Now, the total number of turns is fixed and equal to $$N$$. Therefore the turns per unit length are

$$n=\frac{N}{L}.$$

Substituting this value of $$n$$ into the inductance formula, we obtain

$$L_{\text{ind}} = \mu_0\left(\frac{N}{L}\right)^2 A\,L.$$

We first square the fraction:

$$\left(\frac{N}{L}\right)^2 = \frac{N^2}{L^2}.$$

Putting this back in, we have

$$L_{\text{ind}} = \mu_0\;\frac{N^2}{L^2}\;A\;L.$$

Next we cancel one power of $$L$$ in the denominator with the single $$L$$ in the numerator:

$$L_{\text{ind}} = \mu_0\;N^2\;A\;\frac{1}{L}.$$

All the quantities $$\mu_0,\,N,$$ and $$A$$ are constants under the given conditions, so the inductance is inversely proportional to the length of the solenoid:

$$L_{\text{ind}} \propto \frac{1}{L}.$$

Hence, the correct answer is Option 3.