Answer
43 keV.
$1.4\times10^{-5}$.
Work Step by Step
The Heisenberg uncertainty principle states $\Delta E \Delta t \approx \frac{\hbar}{2}$. Here, $\Delta E$ is the energy width, and $\Delta t$ is the lifetime.
$\Delta E \approx \frac{\hbar}{2 \Delta t}=4.3 \times10^4 eV$
Divide by the rest energy of the $\psi$, which is 3097 MeV (page 1499), to find the fraction of $1.4\times10^{-5}$. The energy width is a tiny fraction of its rest energy.
