A wire carries a steady current 5.0 A along a closed path made of two semicircular arcs of radii 0.10 m and 0.20 m, both centred at the same point O…

Use the Biot–Savart law piece by piece. The two straight segments lie along radial lines through $O$, so $d\vec l\times\hat r=0$ and they contribute nothing. A semicircular arc of radius $\rho$ carrying current $I$ gives, at its centre, half the field of a full loop: $B=\dfrac{\mu_0I}{4\rho}$. The problem states the two arcs' fields add. Hence $B=\dfrac{\mu_0I}{4}\left(\dfrac1{0.10}+\dfrac1{0.20}\right)=\dfrac{(4\pi\times10^{-7})(5)}{4}\times15=1.5708\times10^{-6}\times15=2.356\times10^{-5}\ \mathrm{T}$. That is $23.56\ \mu\mathrm{T}$. A common slip is to use the full-loop formula $\mu_0I/2\rho$, which doubles the result, or to subtract the two fields. Direction matters only for the sign: if the arcs opposed one another, the field would be the difference of the two terms.