= Solution
Restore the unit conversion in the <phase>,
$$
\Delta_{ij}=\frac{\Delta m_{ij}^2c^3L}{4\hbar E}.
$$
Equivalently use <mass> units $\mathrm{eV}/c^2$ and <energies> in $\mathrm{eV}$, with $\hbar c=1.97327\times10^{-7}\,\mathrm{eV\,m}$. The numerical conversion factor is
$$
\boxed{k_{\mathrm{km,MeV,eV}}=\frac{10^3}{4(10^6)(1.97327\times10^{-7})}=1.26693\times10^3.}
$$
Thus
$$
\boxed{\Delta_{ij}=1266.93\,\frac{\Delta m_{ij}^2[\mathrm{eV}^2]L[\mathrm{km}]}{E[\mathrm{MeV}]}.}
$$
The familiar coefficient $1.267$ uses $L$ in metres and $E$ in MeV, or kilometres and GeV; it would be a factor of one thousand too small with the units requested here.<Review of Particle Physics by Particle Data Group (2022)>[Neutrino Masses, Mixing, and Oscillations, equation (14.40): the coefficient 1.267 uses metres per MeV]{ref} The quoted differences with units $\mathrm{eV}^2$ are interpreted as squared-mass differences, despite the source's abbreviated $\Delta m_{ij}$ notation.
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