Given, radius of coil, $a=6 \mathrm{cm}$

$=6\times {10}^{-2}\mathrm{ m}$

Distance, $r=8\mathrm{ cm}$

$=8\times {10}^{-2}\mathrm{ m}$

Magnetic field on the axis of current $\left(I\right)$ carrying coil,

$B_1=\frac{{\mu }_{0}I{a}^2}{2{\left(a^2+r^2\right)}^{3∕2}}$ ...(i)

Magnetic field on the centre of coil

$B_2=\frac{{\mu }_{0}I}{2a}$ ...(ii)

Dividing Equation (ii) by Equation (i), we get

$\frac{B_2}{B_1}=\frac{\frac{{\mu }_{0}I}{2a}}{\frac{{\mu }_{0}I{a}^2}{2{\left(a^2+r^2\right)}^{3/2}}\mathrm{ }}$

$=\frac{{\left(a^2+r^2\right)}^{3/2}}{a^3}$

$=\frac{{\left[{\left(6\times {10}^{-2}\right)}^2+{\left(8\times {10}^{-2}\right)}^2\right]}^{3/2}}{{\left(6\times {10}^{-2}\right)}^3}$

$\Rightarrow \frac{B_2}{B_1}=4.63$

$B_2=4.63\times B_1$

$=4.63\times 54\mathrm{ \mu T}$

$=250\mathrm{ \mu T}$