Explanation

(b) Consider a hypothetical ring of radius $x$ and thickness $kx$ of a disc as shown in figure.
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Charge on the ring, $dq = \frac{q}{\pi r^{2}} \times (2\pi xdx)$
Current due to rotation of charge on ring is
$di = \frac{dq}{T} = \frac{dq}{1/n} = ndq = \frac{nq2xdx}{r^{2}}$
Magnetic field at the centre $O$ due to current of ring element is
$dB = \frac{\mu_{0}di}{2x} = \ \frac{\mu_{0}nq2xdx}{r^{2}(2x)} = \frac{\mu_{0}\ nq\ dx}{r^{2}}$
Total magnetic field induction due to current of whole disc is
$B = \ \int_{0}^{r}{dx = \frac{\mu_{0}\ nq}{r^{2}}(x)_{0}^{2} = \frac{\mu_{0}nq}{r}.}$