The magnetic field at the center of a loop (B₁) is given by

$ B_1 = \frac{\mu_0 I}{2r} $

The magnetic field on the axis of the loop at a distance ( r ) from the center (B₂) is given by

$ B_2 = \frac{\mu_0 Ir^2}{2(r^2 + d^2)^{3/2}} $

where ( d ) is the distance from the center of the coil along the axis. Since ( d = r ), we get

$ B_2 = \frac{\mu_0 I}{4\sqrt{2}r} $

The ratio of $ B_1 $ to $ B_2 $ is

$ \frac{B_1}{B_2} = \frac{\mu_0 I}{2r} \times \frac{4\sqrt{2}r}{\mu_0 I} = \sqrt{8} : 1 $

So, the value of ( x ) is 8.