$\mathrm{B}_{\mathrm{a}}=\frac{\mu_{0} \mathrm{nir}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}$

At the centre,

${\mathrm{x}=0}$

$\therefore $ ${\mathrm{B}_{\mathrm{e}}=\frac{\mu_{0} \mathrm{nir}^{2}}{2 \mathrm{r}^{3}}=\frac{\mu_{0} \mathrm{ni}}{2 \mathrm{r}}}$

$\therefore $ ${\frac{\mathrm{B}_{\mathrm{c}}}{\mathrm{B}_{\mathrm{a}}}=\frac{\mu_{0} \mathrm{ni}}{2 \mathrm{r}} \times \frac{2\left(\mathrm{r}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}{\mu_{0} \mathrm{nir}^{2}}}$

Given : $\quad \mathrm{B}_{\mathrm{c}}=5 \sqrt{5} \mathrm{B}_{\mathrm{a}}$

$\therefore $ $5 \sqrt{5}=\frac{2\left(\mathrm{r}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}{\mathrm{r}^{3}}=\frac{\left(100+\mathrm{x}^{2}\right)^{3 / 2}}{1000}$

Squaring on both sides.

${125=\frac{\left(100+x^{2}\right)^{3}}{10^{6}}} $

or  ${125 \times 10^{6}=\left(100+x^{2}\right)^{3}}$

$100+x^{2}=5 \times 10^{2}=500$

$\therefore \quad \mathrm{x}=20 \,\mathrm{cm}$