Axis of the loop means a line passing through centre of loop and normal to its plane. Since distance of the point P is x from centre of loop and side of square loop is a as shown in Fig. (a).

Therefore, perpendicular distance of P from each side of the loop is$r=\sqrt{x^2+{\left(\frac{a}{2}\right)}^2}=4 \mathrm{cm}$

Now, consider only one side AB of the loop as shown in Fig. (b).

$$\begin{gathered} \tan \alpha =\tan \beta =\frac{(a/2)}{r}=\frac{3}{4} \\ \alpha =\beta ={\tan }^{-1}\left(\frac{3}{4}\right)=37° \\ Magnitude of magnetic induction at P, due to current in this side AB, is \\ {B}_0=\frac{{\mu }_{0}I}{4\pi r}(\sin \alpha +\sin \beta )=3\times {10}^{-6} \mathrm{T} \end{gathered}$$

Components of these magnetic inductions, parallel to plane of loop neutralise each other. Hence, resultant of these two magnetic inductions is $2B_0\cos \theta$ (along the axis).

Similarly, resultantof magnetic inductions produced by currents in remaining two opposite sides BC and AD will also be equal to $2B_0\cos \theta$ (along the axis in same direction). Hence, resultant magnetic induction.

$$\begin{gathered} B=4B_0\cos \theta \\ B=4\times \left(3\times {10}^{-6}\right)\frac{(a/2)}{r}=9\times {10}^{-6} \mathrm{T}=9\mu \mathrm{T} \end{gathered}$$