The magnetic field at the center of the circular loop as shown in figure, when a single wire is bent to form a circular loop and also extends to form straight sections, is

Magnetic field due to ‘AB’;

${\overrightarrow{B}}_1=\frac{{\mu }_{0}I}{4\pi R}\left[\sin \frac{\pi }{2}-\sin \frac{\pi }{4}\right]\otimes =\frac{{\mu }_{0}I}{4\pi R}\left[1-\frac{1}{\sqrt{2}}\right]\otimes$

Magnetic field due to circular loop,  ${\overrightarrow{B}}_2=\frac{{\mu }_{0}I}{2R}\odot$
Magnetic field due to straight wire BC,
$$\begin{gathered} {\overrightarrow{B}}_3=\frac{{\mu }_{0}I}{4\pi R}\left[\sin \frac{\pi }{2}+\sin \frac{\pi }{4}\right]\odot =\frac{{\mu }_{0}I}{4\pi R}\left[1+\frac{1}{\sqrt{2}}\right]\odot \\ \overrightarrow{B}={\overrightarrow{B}}_1+{\overrightarrow{B}}_2+{\overrightarrow{B}}_3 \\ =\left(\frac{{\mu }_{0}I}{2R}+\frac{2{\mu }_{0}I}{4\pi R}\frac{1}{\sqrt{2}}\right)\odot =\frac{{\mu }_{0}I}{2R}\left[1+\frac{1}{\sqrt{2}\pi }\right]\odot \end{gathered}$$