Explanation

(c) Magnetic induction at the centre of the coil of radius $r$ is
$B_{c} = \frac{\mu_{0}nI}{2r}\ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Magnetic induction on the axial line of a circular coil at a distance $x$ from the centre is
$B_{a} = \frac{\mu_{0}nr^{2}I}{2\left( r^{2} + x^{2} \right)^{3/2}}$
Given $x = r$
$\therefore\ \ B_{a} = \frac{\mu_{0}nr^{2}I}{2\left( 2r^{2} \right)^{3/2}}\ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eqs. (i) and (ii), we get
$\frac{B_{c}}{B_{a}} = \frac{2\sqrt{2}}{1}$