Split the wire into parts

The given conductor consists of:

• two straight semi-infinite parts,
• one circular arc of radius $a$ centered at $O$.

We need the magnetic field at the center $O$.

Magnetic field due to the circular arc

For an arc of angle $\theta$ (in radians), the field at the center is $B_{\text{arc}}=\frac{\mu_0 I\theta}{4\pi a}$

From the geometry, the bend is a $\frac{3\pi}{2}$ arc (major arc). Hence, $B_{\text{arc}}=\frac{\mu_0 I}{4\pi a}\cdot \frac{3\pi}{2}$

Magnetic field due to each straight semi-infinite wire

The magnetic field at a perpendicular distance $a$ from a semi-infinite straight wire is $B_{\text{semi-inf}}=\frac{\mu_0 I}{4\pi a}$

There are two such straight parts, so total field from straight portions is $B_{\text{straight}}=2\cdot \frac{\mu_0 I}{4\pi a}=\frac{\mu_0 I}{4\pi a}\cdot 2$

Direction of fields

By the right-hand rule, the field at $O$ due to the arc and both straight portions are in the same direction, so they add.

Net magnetic field

Therefore, $B=B_{\text{arc}}+B_{\text{straight}}$ $B=\frac{\mu_0 I}{4\pi a}\left(\frac{3\pi}{2}+2\right)$

Match with options

This corresponds to Option D.

Comparison with stored answer

Stored correct answer is B: $\frac{\mu_0}{4\pi}\frac{I}{a}\left(\frac{3\pi}{2}+1\right)$

But the standard result for two semi-infinite straight segments gives contribution $2\times \frac{\mu_0 I}{4\pi a}$, not $1\times \frac{\mu_0 I}{4\pi a}$.

So the derived answer is D, not B.