Identify the parts of the wire contributing at the centre $O$ of the circular turn

The wire consists of:

• a long straight part tangent to the circle at $N$,
• a circular arc/turn $DMND$ of radius $R$,
• another long straight part tangent to the circle at $D$.

We need the magnetic field at the centre of the circle.

Magnetic field due to the circular turn

The wire makes one complete circular turn of radius $R$.

Magnetic field at the centre due to a full circle is $B_{\text{circle}}=\frac{\mu_0 I}{2R}$

Writing with denominator $2\pi R$, $B_{\text{circle}}=\frac{\mu_0 I}{2\pi R}\,\pi$

Magnetic field due to the straight tangential parts

For a finite straight wire at perpendicular distance $R$ from the point, $B=\frac{\mu_0 I}{4\pi R}(\sin\theta_1+\sin\theta_2)$

Here each straight part is effectively a semi-infinite wire tangent to the circle.

From the geometry, the centre-to-tangent-point radius is perpendicular to the tangent, and because the two long straight parts meet at right angle, each contributes with angle $45^\circ$ at the far end. Thus field due to each straight part is $B_{\text{one straight}}=\frac{\mu_0 I}{4\pi R}\left(1-\frac{1}{\sqrt2}\right)$ or equivalently, combining the two tangential semi-infinite parts gives a net contribution $B_{\text{straight total}}=\frac{\mu_0 I}{2\pi R}\cdot \frac{1}{\sqrt2}$

Also, by right-hand rule, the field due to the circular loop and the straight sections at the centre are in the same direction, so they add.

Total magnetic field

Therefore, $B=B_{\text{circle}}+B_{\text{straight total}}$ $B=\frac{\mu_0 I}{2\pi R}\pi+\frac{\mu_0 I}{2\pi R}\cdot\frac{1}{\sqrt2}$ $B=\frac{\mu_0 I}{2\pi R}\left(\pi+\frac{1}{\sqrt2}\right)$

Match with the options

This corresponds to: $\boxed{\frac{\mu_0 I}{2\pi R}\left(\pi+\frac{1}{\sqrt2}\right)}$

Hence the correct option is D.