$At \,\,x =\frac{r}{2}$
$B _{ a }=\frac{\mu_{0} Ir ^{2}}{2\left(\frac{ r ^{2}}{4}+ r ^{2}\right)^{3 / 2}}$
$=\frac{\mu_{0} Ir ^{2}}{2\left(\frac{5}{4} r ^{2}\right)^{3 / 2}}=\frac{\mu_{0} I }{2 r }\left(\frac{4}{5}\right)^{3 / 2}$
$=\frac{\mu_{0} I }{2 r }\left(\frac{2}{\sqrt{5}}\right)^{3}$



