A circular coil carrying current has radius ' $R$ '. The distance from the centre of the coil on the axis…

The magnetic induction B at distance $x$ along the axis of a circular coil of radius $R$ carrying current $I$ is given by $B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.

At the coil’s center ($x = 0$), the magnetic field simplifies to $B_c = \frac{\mu_0 I}{2R}$.

Given $B_x = \frac{1}{27} B_c$, substitute the expressions:
$\frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{27} \cdot \frac{\mu_0 I}{2R}$.

Cancel the common factors $\mu_0 I / 2$ from both sides:
$\frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{27R}$.

Cross-multiplying yields $27R^3 = (R^2 + x^2)^{3/2}$.
Raising both sides to the power $\frac{2}{3}$ gives $(27R^3)^{2/3} = (R^2 + x^2)$.

Evaluate the left-hand side:
$(3^3 R^3)^{2/3} = 3^2 R^2 = 9R^2$, so $9R^2 = R^2 + x^2$.

Solving for $x^2$: $x^2 = 8R^2$, hence $x = \sqrt{8R^2} = 2\sqrt{2} R$.

The magnetic induction is $\frac{1}{27}$ its central value at a distance $x = 2\sqrt{2} R$ along the axis.