A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis where the magnetic induction will be `(1//8)^(th)` of its value at the centre of the coil is,

Explanation

Magnetic field induction at a point on the axis at a distance x from the centre of the circular coil carrying current is,
`B_1=(mu_0)/(4pi)(2pinIa^2)/((a^2+x^2)^(3//2))`
Magnetic field induction at the centre of the circular coil carrying current is, `B_2=(mu_0)/(4pi)(2pinI)/(a)`
As per question, `B_1=B_2/8` so `(mu_0)/(4pi)(2pinIa^2)/((a^2+x^2)^(3//2))=(mu_0)/(4pi)(2pinI)/(a)xx1/8` or `(a^2)/((a^2+x^2)^(3//2))=(1)/(8a)`
or `8a^3=(a^2+x^2)^(3//2)` or `2a=(a^2+x^2)^(1//2)` or `4a^2=a^2+x^2` or `x=sqrt3a`