A long straight wire carrying electric current ' i   ' is bent at its mid point to form an…

Recall the formula of the magnetic field due to finite straight wire,

$B=\frac{{\mu }_{0}i}{4\pi r}\left(\sin {\varphi }_1+\sin {\varphi }_2\right)$

here $r=d\cos 45°, {\varphi }_1=90° \& {\varphi }_1=-45°$

$$\begin{gathered} B=\frac{{\mu }_{0}i}{4\pi d\cos 45°}\left(\sin 90°+\sin \left(-45\right)°\right) \\ B=\frac{{\mu }_{0}i}{4\pi d\cos 45°}\left(\sin 90°+\sin \left(-45\right)°\right) \\ B=\frac{{\mu }_{0}i\sqrt{2}}{4\pi d}\left(1-\frac{1}{\sqrt{2}}\right) \\ B=\frac{{\mu }_{0}i}{4\pi d}\left(\sqrt{2}-1\right) \end{gathered}$$