= Solution
Read the original diagram as a clockwise arc carrying <electric current> from its upper endpoint to its lower endpoint, with tangent semi-infinite leads extending to the right. Their crossing is insulated, so there is no extra junction. Choose $+z$ out of the page. At the centre the arc produces a negative axial <magnetic field>, and each straight lead a positive one, by the <right-hand rule>.
For the arc, every element is at distance $R$, with $d\mathbf l$ perpendicular to its displacement to the centre. The <Biot-Savart law> therefore gives
$$
B_{z,\rm arc}=-\frac{\mu_0I}{4\pi R}\int_{-\theta/2}^{\theta/2}d\phi=-\frac{\mu_0I\theta}{4\pi R}.
$$
For each straight lead, write $s\ge0$ for distance along the tangent measured from its endpoint. Its perpendicular distance from the centre is $R$, so
$$
B_{z,\rm lead}=\frac{\mu_0I}{4\pi}\int_0^\infty\frac{R\,ds}{(R^2+s^2)^{3/2}}=\frac{\mu_0I}{4\pi R}.
$$
The integral equals $1/R$ because an <antiderivative> is $s/[R\sqrt{R^2+s^2}]$. Adding both leads and the arc,
$$
B_z(0)=\frac{\mu_0I}{4\pi R}(2-\theta),\qquad\boxed{\theta=2\,\mathrm{rad}\simeq114.6^\circ.}
$$
Using the full infinite-wire <magnetic field> for either lead would incorrectly double its contribution.
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