The magnetic field due to a current carrying circular loop of radius 3 cm at a point on its axis at a distance of 4 cm from the centre is $54\mu T$. The magnetic field (in $\mu T$) at the centre of the loop will be

The magnetic field at a point on the axis of the loop is given by $B=\frac{{\mu }_{0}I{r}^2}{2{\left(r^2+x^2\right)}^{\frac{3}{2}}}\left(1\right)$ 
Magnetic field at the centre of the coil is given by $B=\frac{{\mu }_{0}I}{2r}\left(2\right)$ 
Dividing (1) by (2), we get
$\frac{B_0}{B}=\frac{r^3}{{\left(r^2+x^2\right)}^{\frac{3}{2}}}$ 
Substituting the values of r and x, we get
$\begin{array}{l}\frac{B_0}{B}=\frac{125}{27}\text{\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}or}\\ B_0=\frac{125}{27}B=\frac{125}{27}\times 54\mu T=250\mu T\end{array}$ 
Hence the correct choice is (1)