Magnetic field due to a straight wire will be zero if the line of the wire passes through the point of interest. Therefore, In the above figure,

$B_{BC}+B_{DE}+B_{FG}=0$

Therefore,

${\overrightarrow{B}}_{\mathrm{net} }={\overrightarrow{B}}_{AB}+{\overrightarrow{B}}_{CD }+{\overrightarrow{B}}_{EF}$

Magnetic field due to a finite wire is given by,

$B=\frac{{\mu }_{0}I}{4\pi d}\left(\sin \alpha +\sin \beta \right)$

Using above formula,

${\overrightarrow{B}}_{AB}=\frac{{\mu }_{0}I}{4\pi L}\times \frac{1}{\sqrt{2}}\left(-\hat{\mathrm{k}}\right)$

Magnetic field due to a circular arc is,

$B=\frac{{\mu }_{0}I}{2R}\times \frac{\theta }{2\pi }$

Therefore,

${\overrightarrow{B}}_{CD}=\frac{{\mu }_{0}I}{4\left(\frac{L}{2}\right)}\left(-\hat{\mathrm{k}}\right)$ and ${\overrightarrow{B}}_{EF}=\frac{{\mu }_{0}I}{8\times \left(\frac{L}{4}\right)}\left(-\hat{\mathrm{k}}\right)$

Therefore, net magnetic field

${\overrightarrow{B}}_{\mathrm{net}}=\frac{{\mu }_{0}I}{4\pi \times L}\left(\frac{1}{\sqrt{2}}\right)+\frac{{\mu }_{0}I}{4\left(\frac{L}{2}\right)}+\frac{{\mu }_{0}I}{8\times \left(\frac{L}{4}\right)}\left(-\hat{\mathrm{k}}\right)$

$\Rightarrow {\overrightarrow{B}}_{\mathrm{net}}=\left(\frac{{\mu }_{0}I}{2L}+\frac{{\mu }_{0}I}{2L}+\frac{{\mu }_{0}I}{4\sqrt{2}\pi L}\right)\left(-\hat{\mathrm{k}}\right)$

$\Rightarrow {\overrightarrow{B}}_{\mathrm{net}}=\frac{{\mu }_{0}I}{L}\left(1+\frac{1}{4\sqrt{2}\pi }\right)\left(-\hat{\mathrm{k}}\right)$