Let the solenoid of figure consists of $n$ turns per unit length.

Let its length be $2 l$ and radius $a$. We can evaluate the axial field at a point $P$, at a distance $r$ from the centre $\mathrm{O}$ of the solenoid.

Consider a circular element of thickness $d x$ of the solenoid at a distance $x$ from the centre. It consists of $n d x$ turns. Let $I$ be the current in the solenoid. According to formula for the magnitude of magnetic field at axis of coil of $\mathrm{N}$ turn, number of turn $\mathrm{N}=n d x$ and taking distance from $\mathrm{O}$ to $\mathrm{P}=(r-x)$.

$\mathrm{B}=\frac{\mu_{0} \mathrm{NI} a^{2}}{2\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}$

$\therefore \mathrm{B}=\frac{\mu_{0} n d x \mathrm{I} a^{2}}{2\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}$

Total field is obtained by summing over all the elements that is by integrating from $x=-l$ to $x=+l$. Then the denominator is approximately by,

$\left[(r-x)^{2}+a^{2}\right]^{3 / 2} \approx r^{3} \quad[\because l$ means $x$ and $a$ is neglected to compare with $r]$

$\therefore \mathrm{B}=\frac{\mu_{0} n \mathrm{I} a^{2}}{2 r^{3}} \int_{-l}^{l} d x$

$=\frac{\mu_{0} n \mathrm{I} a^{2}}{2 r^{3}}[l-(-l)]$