To find the total magnetic field at point $O$, we consider the three segments of the wire independently and use the principle of superposition:

1. Semi-infinite straight wires: There are two very long (semi-infinite) linear segments parallel to the X-axis. According to the sources, the magnetic field due to a semi-infinite wire at a distance $R$ from its end is $B = \frac{\mu_0 I}{4\pi R}$ . Applying the right-hand thumb rule for the given geometry, both segments produce a field at $O$ in the negative Z-direction ($-\hat{k}$). Total field from linear parts = $2 \times \left(\frac{\mu_0 I}{4\pi R}\right)(-\hat{k}) = -\frac{2\mu_0 I}{4\pi R}\hat{k}$.

2. Semicircular arc: For a circular arc of radius $R$ subtending an angle $\theta$ at the center, the field is $B = \frac{\mu_0 I \theta}{4\pi R}$ . For a semicircle, $\theta = \pi$, so $B = \frac{\mu_0 I \pi}{4\pi R}$. Since the arc lies in the Y-Z plane, its axial field at the center $O$ will be along the X-axis. Given the current direction, the field points in the negative X-direction ($-\hat{i}$). Field from arc = $-\frac{\pi\mu_0 I}{4\pi R}\hat{i}$.

3. Total Field: Summing the vectors: $\vec{B} = -\frac{\pi\mu_0 I}{4\pi R}\hat{i} - \frac{2\mu_0 I}{4\pi R}\hat{k} = -\frac{\mu_0 I}{4\pi R}(\pi\hat{i} + 2\hat{k})$. This matches Option 3.