If the current 'i' is passing through the wire, it has a shape as...

The wire can be divided into three parts: a lower semi-infinite straight wire, a semicircular arc, and an upper semi-infinite straight wire.

The magnetic field at point O due to the lower semi-infinite straight wire is given by $B_1 = \dfrac{\mu_0 i}{4\pi R}$. Applying the right-hand rule, the current is towards the left and point O is above the wire, so the magnetic field is directed into the page.

The magnetic field at point O due to the semicircular arc is given by $B_2 = \dfrac{\mu_0 i}{4R} = \dfrac{\pi \mu_0 i}{4\pi R}$. The current is in the clockwise direction, so the magnetic field is directed into the page.

The magnetic field at point O due to the upper semi-infinite straight wire is given by $B_3 = \dfrac{\mu_0 i}{4\pi R}$. The current is towards the right and point O is below the wire, so the magnetic field is directed into the page.

Since all three magnetic field components are in the same direction, the total magnetic field at O is their sum:

$B_{net} = B_1 + B_2 + B_3$

$B_{net} = \dfrac{\mu_0 i}{4\pi R} + \dfrac{\pi \mu_0 i}{4\pi R} + \dfrac{\mu_0 i}{4\pi R}$

$B_{net} = \dfrac{\mu_0 i}{4\pi R}(1 + \pi + 1)$

$B_{net} = \dfrac{\mu_0 i}{4\pi R}(2 + \pi)$