The magnitude of the magnetic field at point O (centre of the arc)...

The magnetic field at the center O is the sum of the magnetic fields due to the straight wire segments and the circular arc.

The straight wire segments lie along the radial lines passing through the center O. For these segments, the angle between the current element $d\vec{l}$ and the position vector $\vec{r}$ is 0^{}^{\circ} or 180^{}^{\circ}. Thus, $d\vec{l} \times \vec{r} = 0$, and their contribution to the magnetic field at O is zero.

The circular arc subtends an angle \theta = 360^{}^{\circ} - 90^{}^{\circ} = 270^{}^{\circ} = \dfrac{3\pi}{2} radians at the center.

The magnetic field due to a circular arc at its center is given by:

$B = \dfrac{\mu_0 I}{4\pi R} \theta$

Substituting $\theta = \dfrac{3\pi}{2}$:

$B = \dfrac{\mu_0 I}{4\pi R} \left(\dfrac{3\pi}{2}\right) = \dfrac{3\mu_0 I}{8R}$

Answer: $\dfrac{3\mu_0 I}{8R}$