= Solution
Assume coherent production and detection, relativistic <neutrinos>, two active <neutrino flavours>, propagation in vacuum, negligible packet separation and no absorption. Invert the given real <lepton mixing matrix> to obtain $|\nu_\mu\rangle=c|\nu_2\rangle-s|\nu_3\rangle$ and $|\nu_\tau\rangle=s|\nu_2\rangle+c|\nu_3\rangle$, where $c=\cos\theta$, $s=\sin\theta$. With phases $\phi_a=E_at-\mathbf p_a\cdot\mathbf x$ in <natural units>,
$$
|\nu(L)\rangle=c e^{-i\phi_2}|\nu_2\rangle-s e^{-i\phi_3}|\nu_3\rangle,\qquad
\langle\nu_\tau|\nu(L)\rangle=cs(e^{-i\phi_2}-e^{-i\phi_3}).
$$
Using the paper's opposite sign convention conjugates this <probability amplitude> and leaves the <probability> unchanged. For equal <momentum>, $E_a\simeq p+m_a^2/(2p)$ and $t\simeq L$, so $\Delta\phi=\Delta m^2L/(2E)$. The equal-<energy> convention gives the same leading result from the <momentum> difference; evaluating each component at a different arrival time would incorrectly double the phase.
$$
P_{\mu\tau}=4c^2s^2\sin^2\frac{\Delta\phi}{2}=\sin^2(2\theta)\sin^2\frac{\Delta m^2L}{4E},\qquad P_{\mu\mu}=1-P_{\mu\tau}.
$$
Since $\hbar c=1.97327\times10^{-7}\,\mathrm{eV\,m}$,
$$
\frac{\Delta m^2L}{4E\hbar c}=1.26693\frac{\Delta m^2[\mathrm{eV}^2]L[\mathrm{km}]}{E[\mathrm{GeV}]}.
$$
This gives the stated factor $1.27$. <Neutrino oscillation> in this approximation measures $|\Delta m^2|$, not the absolute <neutrino mass> or its ordering.
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