= Solution
At the largest admitted orbit, $r=R$ and $m_pv_{\max}=eBR$. Consequently
$$
\boxed{v_{\max}=\frac{eBR}{m_p},\qquad T_{\max}=\frac{e^2B^2R^2}{2m_p}.}
$$
Using $e=1.602176634\times10^{-19}\,\mathrm C$ and $m_p\simeq1.67262\times10^{-27}\,\mathrm{kg}$ with the specified $B$ and $R$ yields
$$
\boxed{T_{\max}\simeq1.92\times10^{-12}\,\mathrm J\simeq11.97\,\mathrm{MeV},\qquad v_{\max}\simeq4.79\times10^7\,\mathrm{m\,s^{-1}}.}
$$
This is about $0.160c$; the displayed results use the nonrelativistic approximation explicitly requested by the paper. The drive amplitude controls the energy gain per gap crossing and hence the number of turns, but the ideal orbit-radius limit is set by $B$ and $R$.
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