$\mathrm{d} \mathrm{N}=\left(\frac{\mathrm{N}}{\mathrm{b}-\mathrm{a}}\right) \mathrm{dr}$

Magnetic field due to this element at the centre of the centre of the coil will be

$\mathrm{dB}=\frac{\mu_{0}(\mathrm{dN}) \mathrm{I}}{2 \mathrm{R}}=\frac{\mu_{0} \mathrm{I}}{2} \frac{\mathrm{N}}{\mathrm{b}-\mathrm{a}} \cdot \frac{\mathrm{dr}}{\mathrm{r}}$

$\therefore B=\int_{r=a}^{r-b} d B=\frac{\mu_{0} N I}{2(b-a)} \operatorname{In}\left(\frac{b}{a}\right)$
The idea of the question is taken from question number $3.245$ of $I.E.$ lrodov.