Here, $a =\frac{r}{\sqrt{2}}$
Magnetic field at point O due to AB is
$B_{1}=\frac{\mu_{0}}{4\pi } \frac{I}{a}=\frac{\mu_{0} I}{4\pi\left(r /\sqrt{2}\right)}$
Magnetic field at point O due to BCD is
$B_{2}=\frac{\mu_{0}I}{4\pi r} \left(\frac{\pi}{2}\right)$
Magnetic field at point O due to DE is
$B_{3}=\frac{\mu_{0}I}{4\pi a}=\frac{\mu_{0}I}{4\pi\left(r /\sqrt{2}\right)}$
Resultant magnetic field at point O is
$B = B_{1} + B_{2} + B_{3}$
$=\frac{\mu_{0}I}{4\pi\left(r /\sqrt{2}\right)}+\frac{\mu_{0}I}{4\pi r} \left(\frac{\pi}{2}\right)+\frac{\mu_{0}I}{4\pi\left(r /\sqrt{2}\right)}$
$=\frac{\mu_{0}I}{4\pi r}\left(\sqrt{2}+\frac{\pi}{2}+\sqrt{2}\right)=\frac{\mu_{0}I}{4\pi r}\left(2\sqrt{2}+\frac{\pi}{2}\right)$
$=\frac{\mu_{0}2I}{4\pi r}\left(\sqrt{2}+\frac{\pi}{4}\right)$
$=\frac{10^{-7}\times2I}{r}\left(\sqrt{2}+\frac{\pi}{4}\right)\quad\quad\quad\left[\because \frac{\mu_{0}}{4\pi}=10^{-7}\right]$
