= Solution
Interpret the percentage relative to the classical <kinetic energy>, so that $K_{\rm rel}=1.02K_{\rm cl}$. Set $\beta=v/c$ and use the <Lorentz factor> $\gamma=(1-\beta^2)^{-1/2}$. The <relativistic kinetic energy> and classical <kinetic energy> give
$$
(\gamma-1)mc^2=1.02\left(\frac12mv^2\right),
\qquad \gamma-1=0.51\beta^2.
$$
For the positive nonzero <speed>, insert $\beta^2=1-\gamma^{-2}$ and cancel $\gamma-1$:
$$
\frac{\gamma^2}{\gamma+1}=0.51,
\qquad \gamma^2-0.51\gamma-0.51=0,
\qquad \gamma=\frac{0.51+\sqrt{2.3001}}2\simeq1.01330403.
$$
It follows that
$$
\boxed{v=c\sqrt{1-\gamma^{-2}}\simeq0.161513c
\simeq4.8420\times10^7\,\mathrm{m\,s^{-1}},}
\qquad
\boxed{E=\gamma mc^2\simeq950.479\,\mathrm{MeV}.}
$$
Here $mc^2=938\,\mathrm{MeV}$ is the <proton>'s <rest energy>; the resulting <relativistic kinetic energy> is $12.479\,\mathrm{MeV}$. The zero-<speed> equality before cancellation has no meaningful percentage excess because both <kinetic energies> then vanish.
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