Answer
$$E_{binding} = 11.161\times 10^{-16} J = 6.966KeV$$
Work Step by Step
Since the electron ejected moves at a speed closer to the speed of light. We should find Kinetic energy of electron based on the relativistic mass to get accurate results.
$$m = \frac{m_e}{\sqrt{1-(\frac{v}{c})^2}} =\frac{9.109\times 10^{-31}}{\sqrt{1-\big(\frac{21.4\times 10^6}{3\times 10^8}\big)^2}} = 9.133\times 10^{-31}kg$$
$$E_{photon} = E_{binding} + E_{K.E}$$
$$E_{binding} = E_{photon} - E_{K.E}$$
$$E_{binding} = \frac{hc}{\lambda} - \frac{1}{2}mv^2$$
$E_{binding} = \frac{6.626\times 10^{-34} \times 3\times 10^8}{150\times 10^{-12}} - \frac{1}{2}(9.133\times 10^{-31})(21.4\times 10^6)^2$
$$E_{binding} = 11.161\times 10^{-16} J = 6.966KeV$$
