A wire carrying current I forms a closed loop made of two circular arcs, each subtending an angle of 120^° at the common centre O, of radii R and 3R…

Radial segments pass through $O$, so $d\vec l\times\hat r=0$ and they contribute nothing. For a circular arc of radius $a$ subtending $\theta$ at its centre, the field at the centre is $B=\dfrac{\mu_0I}{4\pi}\dfrac{\theta}{a}$ (from $B=\int\dfrac{\mu_0}{4\pi}\dfrac{I\,a\,d\phi}{a^2}$). Here $\theta=\dfrac{2\pi}{3}$, so $\dfrac{\mu_0I\theta}{4\pi}=\dfrac{\mu_0I}{6}$. The inner arc gives $\dfrac{\mu_0I}{6R}$ and the outer arc gives $\dfrac{\mu_0I}{18R}$. Their senses of circulation are opposite (inner anticlockwise, outer clockwise), so the fields at $O$ are antiparallel and the net field is $\dfrac{\mu_0I}{6R}-\dfrac{\mu_0I}{18R}=\dfrac{\mu_0I}{9R}$. Adding the magnitudes would give $\dfrac{2\mu_0I}{9R}$, and using only one arc gives $\dfrac{\mu_0I}{6R}$ or $\dfrac{\mu_0I}{18R}$.