Explanation

(a) Magnetic field at the centre of a circular loop of radius $R$ carrying current $I$ is $B = \frac{\mu_{0}2\pi I}{4\pi R} = \frac{\mu_{0}I}{2r}$
Its magnetic moment is $M = IA = I(\pi R^{2})$
$\therefore\frac{B}{M} = \frac{\mu_{0}I}{2R} \times \frac{1}{I\pi R^{2}} = \frac{\mu_{0}}{2\pi R^{2}} = x\ \ \ \lbrack Given\rbrack$
When both the current and radius is doubled, the ratio becomes $\frac{B^{'}}{M'} = \frac{\mu_{0}}{2\pi(2R)^{3}} = \frac{1}{8}\left( \frac{\mu_{0}}{2\pi R^{3}} \right) = \frac{x}{8}$