Hint 7

What is the square of the 4-momentum of the photon? What is the minimum magnitude for the square of the total 4-momentum of the electron-positron pair?

Use 4-momentum conservation and work with the square of the total 4-momentum. The minimum energy is required when the final particles are all at rest relative to each other and so the square of the total 4-momentum is $-(3m_ec)^2$ in any inertial frame. In the frame of interest the 4-momenta of the photon and electron are, respectively,

$$(E_\gamma/c)(1,0,0,i)\quad\hbox{and}\quad(0,0,0,im_ec).$$

Squaring their sum gives

$$-m_e^2c^2-2m_eE_\gamma=-9m_e^2c^2,$$

from which $E_\gamma=4m_ec^2=2.04$ MeV follows.