Magnetic field at centre: $B_0 = \frac{\mu_0 I}{2r}$

Magnetic field on axis at distance $x$:
$$B_{axis} = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$$

Given $B_{axis} = B_0/8$:
$$\frac{r^2}{(r^2 + x^2)^{3/2}} = \frac{1}{8} \cdot \frac{1}{r}$$
$$(r^2 + x^2)^{3/2} = 8r^3$$
$$r^2 + x^2 = 4r^2 \Rightarrow x^2 = 3r^2$$
$$x = \sqrt{3}\, r$$