= Solution
At time $t_0$, the accelerated <proton> has <momentum> $p=qEt_0$ and, by the <relativistic energy-momentum relation>, energy
$$
\mathcal E_p=\sqrt{m^2c^4+q^2E^2t_0^2c^2}.
$$
The incident <cosmic microwave background> <photon> adds energy $E_\gamma$. Since the final state contains only the <Delta-plus baryon>, <conservation of relativistic energy> gives its laboratory-frame <Lorentz factor>
$$
\gamma_\Delta=\frac{\mathcal E_p+E_\gamma}{m_\Delta c^2}.
$$
Its rest-frame decay time $t_\Delta$ is a <proper time>, so <time dilation> gives the laboratory decay time
$$
\boxed{t_{\rm lab}=\frac{t_\Delta}{m_\Delta c^2}\left(\sqrt{m^2c^4+q^2E^2t_0^2c^2}+E_\gamma\right)}.
$$
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