The magnetic field at O due to the wire loop ABCD carrying a current I as shown in figure is

 

 

Magnetic field at O due to straight segments BC and AD is zero. For the curved part AB, $\theta =\frac{3\pi }{2}$ Therefore,

$B_{AB}=\frac{3{\mu }_{0}I}{8 r}$  directed into the page.

For the curved part CD, $\theta =\frac{\pi }{2}.$ Therefore,

$B_{CD}=\frac{{\mu }_{0}I}{2\left(2r\right)}x\frac{{\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2\mathrm{\pi }}=\frac{{\mu }_{0}I}{16r}$  directed into the page.

Therefore, magnetic field at O due to current I in ABCD is

$$\begin{gathered} B=B_{AB}+B_{CD} \\ =\frac{3{\mu }_{0}I}{8r}+\frac{{\mu }_{0}I}{16r}=\frac{7{\mu }_{0}I}{16r} directed into the page. \end{gathered}$$