= Solution
Restoring units in the <oscillation phase> gives $\Delta_{ij}=\Delta m_{ij}^2L/(4E\hbar c)$. With $\hbar c=1.97327\times10^{-7}\ \mathrm{eV\,m}$, one kilometre is $5.06773\times10^9\ \mathrm{eV}^{-1}$. Thus
$$
\Delta_{ij}=\frac{5.06773\times10^9}{4\times10^6}
\frac{\Delta m_{ij}^2[\mathrm{eV}^2]L[\mathrm{km}]}{E[\mathrm{MeV}]},
\qquad \boxed{k_{\rm practical}=1.26693\times10^3.}
$$
Here $k_{\rm practical}$ denotes the coefficient in $\Delta=k_{\rm practical}\Delta m^2[\mathrm{eV}^2]L[\mathrm{km}]/E[\mathrm{MeV}]$. The familiar $1.267$ applies when $E$ is in GeV, not MeV. Strictly, the dimensionless definition $\Delta=2k(\phi_i-\phi_j)$ keeps $k=1/4$ once the $\phi_j$ already include the unit conversion. The intended numerical question concerns the practical coefficient multiplying the numerical values of $\Delta m^2L/E$.
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