A wire loop carrying current I is placed in x-y plane. The magnetic field at O is

Due to straight wire 

$B_1=\frac{{\mu }_{0}I}{4\pi \cdot r\cos {60}^0}(\cos {30}^0+\cos {30}^0)=\frac{{\mu }_{0}I}{2\pi r}\cdot \sqrt{3},\otimes$

Due to circular portion

$B_2=\frac{{\mu }_{0}I}{2R}\cdot \frac{{120}^0}{{360}^0}=\frac{{\mu }_{0}I}{6R},\odot$

$B_1>B_2$

$B_0=\frac{{\mu }_{0}I}{2r}[\frac{\sqrt{3}}{\pi }-\frac{1}{3}],\otimes$