A closed wire loop consists of two concentric semicircular arcs of radii 0.10 m and 0.20 m having the common centre O, joined by two straight radial…

The radial straight segments pass through $O$ (their lines contain $O$), so they contribute no field there. Each semicircular arc of radius $r$ gives $B=\frac{\mu_0I}{4r}$. Going round the loop, the current traverses the two arcs in opposite senses (one clockwise, the other anticlockwise), so their fields at $O$ are opposite: $B=\frac{\mu_0I}{4}\left(\frac1{0.10}-\frac1{0.20}\right)=\frac{4\pi\times10^{-7}\times10}{4}\times5=1.5708\times10^{-5}\ \mathrm{T}$. The field at $O$ is normal to the plane of the loop, while the proton moves in that plane, so $\vec v\perp\vec B$ and the force is $F=evB=1.6\times10^{-19}\times2\times10^{5}\times1.5708\times10^{-5}=5.03\times10^{-19}$ N. Hence $x=5.03$. Treating both arcs as adding would give $47.1\ \mu\mathrm{T}$ and a force three times larger.