1. Interpret the geometry

The wire is infinitely long, and its cross-section is a semi-circular ring of radius $R$.

This means the conductor consists of many thin long filaments parallel to the axis, distributed uniformly over a semicircular arc of radius $R$.

We need the magnetic field along the axis of the wire, i.e. at the center of this semicircular cross-section.

2. Magnetic field due to one thin filament

For an infinitely long straight wire carrying current $dI$ at distance $R$ from the axis point,

$dB = \frac{\mu_0 \, dI}{2\pi R}$

The direction is tangential around that filament. At the center of the semicircle, only the components perpendicular to the symmetry axis cancel appropriately, while the components along the diameter-bisector add.

Let the filament be at angular position $\theta$ on the semicircle. Then the contributing component is

$dB_{\text{net}} = dB \cos\theta$

where $\theta$ is measured from the symmetry axis of the semicircle.

So,

$dB_{\text{net}} = \frac{\mu_0 \, dI}{2\pi R}\cos\theta$

3. Current distribution over the semicircular ring

Since the current is uniformly distributed over the semicircular arc, current per unit length of arc is

$\lambda = \frac{I}{\pi R}$

A small arc element is

$dl = R\, d\theta$

Hence current in this element is

$dI = \lambda \, dl = \frac{I}{\pi R}(R\,d\theta)=\frac{I}{\pi}d\theta$

4. Integrate over the semicircle

Take the semicircle from $-\pi/2$ to $\pi/2$.

Then

$B = \int dB_{\text{net}} = \int_{-\pi/2}^{\pi/2} \frac{\mu_0}{2\pi R}\cdot \frac{I}{\pi}\cos\theta\, d\theta$

$B = \frac{\mu_0 I}{2\pi^2 R} \int_{-\pi/2}^{\pi/2} \cos\theta\, d\theta$

Now,

$\int_{-\pi/2}^{\pi/2} \cos\theta\, d\theta = \left[\sin\theta\right]_{-\pi/2}^{\pi/2}=1-(-1)=2$

Therefore,

$B = \frac{\mu_0 I}{2\pi^2 R}\cdot 2 = \frac{\mu_0 I}{\pi^2 R}$

5. Match with options

Thus the magnetic induction along the axis is

$\boxed{\frac{\mu_0 I}{\pi^2 R}}$

This matches Option D.

6. Comparison with stored answer

Stored correct answer: D

Derived answer: D

So, the derived answer agrees with the stored answer.