Number of turns in this elementary ring $dN =\frac{ N }{ b - a } dr$

Thus magnetic field at the centre $O$ due to this ring $dB =\frac{\mu_0 IdN }{2 r }$

We get $dB =\frac{\mu_0 INdr }{2(b- a ) r }$

Net magnetic field at centre of spiral $B=\int_a^b \frac{\mu_0 I N ~ d r}{2(b-a) r}$

$\therefore B =\frac{\mu_0 IN }{2(b- a )} \int_{ a }^{ b } \frac{ dr }{ r }$

Or $B=\frac{\mu_0 I N}{2(b-a)} \times\left.\ln r\right|_a ^b$

Or $B=\frac{\mu_0 I N}{2(b-a)} \ln \frac{b}{a}$