= Solution
The threshold must conserve both <energy> and <momentum>. For head-on <photons>, the total <energy> is $E_1+E_2$, while the total <momentum> has magnitude $|E_2-E_1|/c$. Their combined <invariant mass> $M$ satisfies
$$
M^2c^4=(E_1+E_2)^2-c^2\left(\frac{E_2-E_1}{c}\right)^2=4E_1E_2.
$$
At the threshold for <photon-photon pair production>, the <electron> and <Positron> have no relative <momentum> in their <centre-of-momentum frame>, so $M=2m_e$. Therefore
$$
\boxed{E_{2,\min}=\frac{(m_ec^2)^2}{E_1}=\frac{(5.110\times10^5\ \mathrm{eV})^2}{10^{-3}\ \mathrm{eV}}\simeq2.61\times10^{14}\ \mathrm{eV}=261\ \mathrm{TeV}.}
$$
Simply equating the laboratory <energy> to $2m_ec^2$ would miss the very large laboratory <momentum>: the newly produced pair must carry it away.
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