Answer
${\bf 1.26\times 10^{-6}}\;\rm T\cdot m$
Work Step by Step
We have here 3 parts of the path, from point $i$ to the left side of the semicircle, the semicircle itself, and from the right side of the semicircle to point $f$.
According to Ampere's law, the line integral of the magnetic field is given by
$$\int_i^f \vec B\cdot d\vec s=\int_{\rm left} \vec B\cdot d\vec s +\int_{\rm semicircle} \vec B\cdot d\vec s +\int_{\rm right} \vec B\cdot d\vec s $$
We can see that the magnetic field if the wire is always perpendicular to the two horizontal lines while it is parallel to the semicircle and clockwise which means in the same direction of the path.
So,
$$\int_i^f \vec B\cdot d\vec s= B s_{\rm left}\cos90^\circ + B (\pi r) +B s_{\rm right}\cos90^\circ $$
$$\int_i^f \vec B\cdot d\vec s= 0 + \dfrac{\mu_0I}{2\pi d} (\pi r) +0 $$
where $r=d$,
$$\int_i^f \vec B\cdot d\vec s= \dfrac{\mu_0I}{2 } $$
Plug the known;
$$\int_i^f \vec B\cdot d\vec s= \dfrac{(4\pi \times 10^{-7})(2)}{2 } $$
$$\int_i^f \vec B\cdot d\vec s=\color{red}{\bf 1.26\times 10^{-6}}\;\rm T\cdot m$$
