Solution

The problem can be solved simply by writing down the conservation of momentum and energy, but I want to show another method. One can notice that if we go to a reference frame where the electrons in the beam are at rest, we observe the Compton effect. The change in energy upon scattering of a photon in this case can be obtained from the well-known formula

$$\frac{1}{\nu'}-\frac{1}{\nu}=\frac{h}{m_e c^2}(1-\cos\alpha)$$
where $\alpha$ is the scattering angle of the photon. This formula is derived in problem $14.5.23$.

As a result, we need to carefully transform to such a system, calculate the energy, and then transform back to the laboratory frame.

For convenience, we switch to a system of units where $c = 1$. This simplifies the notation, and it is convenient that the photon momentum equals its energy. We will denote energies in the laboratory frame (LF) by $\mathcal{E}$, and in the electron rest frame (RF) by $E$.

Thus, the gamma factor for the electron in the LF is obviously
\gamma=(1-v^2)^{-\frac{1}{2}}=\frac{ \mathcal{E}_e }{m_e}, \ \to \ v=\sqrt{1-\left(\frac{m_e}{ \mathcal{E}_e }\right)^2}=\frac{\sqrt{\mathcal{E}_e^2-m_e^2}}{\mathcal{E}_e}
$$This is the speed at which the RF moves relative to the LF. We find the photon energy in the RF using the relativistic Doppler effect:$$
E=\mathcal{E}\gamma(1+v)
$$

Using the Compton formula, we obtain the energy change upon collision:
$$E'=\frac{Em_e}{E(1-\cos\alpha)+m_e}$$

The maximum energy in the LF is achieved when the photon is scattered backward in the RF ($\alpha = \pi$), because upon inverse transformation to the LF, a photon scattered backward in the RF receives the maximum Doppler shift forward (in the direction of the electron's motion):
$$E'=\frac{Em_e}{2E+m_e}$$
Inverse transformation (taking into account that the photon's direction has reversed):
$$\mathcal{E}'=E'\gamma(1+v)$$
$$\mathcal{E}'_{max} =\frac{Em_e}{2E+m_e}\gamma(1+v)=\frac{\mathcal{E}\gamma^2(1+v)^2m_e}{2\mathcal{E}\gamma(1+v)+m_e}$$
Multiply numerator and denominator by $(1 - v)$:
$$\mathcal{E}'_{max} =\frac{\mathcal{E}m_e(1+v)}{2\mathcal{E}\gamma(1-v^2)+m_e(1-v)}$$
$$\mathcal{E}'_{max} =\frac{\mathcal{E}m_e(1+v)\gamma}{2\mathcal{E}+m_e\gamma(1-v)}$$
$$\mathcal{E}'_{max} =\mathcal{E}\frac{\mathcal{E}_e(1+v)}{2\mathcal{E}+\mathcal{E}_e(1-v)}$$
Finally substitute the velocity:
$$\mathcal{E}'_{max} =\mathcal{E}\frac{\mathcal{E}_e+\sqrt{\mathcal{E}_e^2-m_e^2}}{2\mathcal{E}+\mathcal{E}_e-\sqrt{\mathcal{E}_e^2-m_e^2}}$$

Now return to standard units:

$$\mathcal{E}'_{max} =\mathcal{E}\frac{\mathcal{E}_e+\sqrt{\mathcal{E}_e^2-(m_e c^2)^2}}{2\mathcal{E}+\mathcal{E}_e-\sqrt{\mathcal{E}_e^2-(m_e c^2)^2}}$$
Don't forget the numerical value (Savchenko forgot...):
$$\mathcal{E}_{max}\approx 6\cdot10^9 \ эВ$$
Note: Savchenko apparently has a small typo in the answer.