Pls help me to solve 47

Solution,
Magnetic field at a point on the axis of the circular coil carrying current is given by:
$$\begin{gathered} B=\frac{{\mu }_{o}nI{a}^2}{2{(a^2+x^2)}^{3/2}} \\ where, \\ n=no. of turns \\ I=current through the coil \\ a=radius of the coil \\ x=dis\tan ce of the point from the centre of the coil. \end{gathered}$$

Given,
I = 5 A
n = 500
a = 5cm = 0.05 m
x = 12 cm = 0.12 m
Then,
$B=\frac{{\mu }_{o}nI{a}^2}{2{(a^2+x^2)}^{3/2}}=\frac{4\pi \times {10}^{-7}\times 500\times 5\times 0.{05}^2}{2{(0.{05}^2+0.{12}^2)}^{3/2}}=1.786 \times {10}^{-3}T = 17.9 G = 18 G$