A long wire carrying a current I=5.0 A is bent into a hairpin: two straight parallel semi-infinite sections, separated by 2R with R=10 cm, are joined…

Take the current to enter along one straight section, traverse the semicircle and leave along the other. For the arc, the Biot-Savart law gives $B_{\text{arc}}=\dfrac{\mu_0I}{4R}$ at the centre. Each semi-infinite straight wire ends at the foot of the perpendicular from $O$, so $B_{\text{wire}}=\dfrac{\mu_0I}{4\pi R}$. By the right-hand rule all three contributions point in the same direction (perpendicular to the plane), so $B=\dfrac{\mu_0I}{4R}\left(1+\dfrac2\pi\right)$. Numerically, $\dfrac{\mu_0I}{4R}=\dfrac{4\pi\times10^{-7}\times5}{0.4}=15.71\ \mu\mathrm{T}$, and multiplying by $1.6366$ gives $25.71\ \mu\mathrm{T}$. The semi-infinite result follows from the finite-wire formula with angle limits $0$ and $90^\circ$, which is why each straight arm gives half of the infinite-wire value.