= Solution
Let the incident <photon energies> be $E$ and $\varepsilon$, meeting at angle $\psi$ in the <laboratory frame>. Add their <four-momenta>; because each is null, the squared total invariant <energy> is
$$
\mathcal S=(E+\varepsilon)^2-c^2|\mathbf p_1+\mathbf p_2|^2
=2E\varepsilon(1-\cos\psi).
$$
For <photon-photon pair production>, its threshold value is $(2m_ec^2)^2$: the <electron> and <Positron> have zero relative <momentum> in their <centre-of-momentum frame>. The minimum possible $E$ occurs for a head-on collision, $\psi=\pi$, so
$$
4E\varepsilon=4m_e^2c^4,
\qquad\boxed{E_{\min}=\frac{(m_ec^2)^2}{\varepsilon}.}
$$
With $m_ec^2\simeq0.511\,\mathrm{MeV}$ and $\varepsilon=10^{-3}\,\mathrm{eV}$,
$$
\boxed{E_{\min}\simeq2.61\times10^{14}\,\mathrm{eV}
=261\,\mathrm{TeV}\simeq41.8\,\mathrm J.}
$$
The quoted background <photon energy> is used directly, without substituting a different characteristic value of the <cosmic microwave background>. A less opposed collision has a higher threshold; parallel <photons> cannot produce this massive pair.
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