Solution

The initial velocity is zero, therefore
$$v=\int_0^t a \, dt \tag{1}$$
Newton's second law:
$$F=-Ee=am_e \tag{2}$$
Let $a_0=\frac{eE_0}{m_e}$, then from (2) into (1):
$$v(t)=-a_0\int_0^t \sin (\omega t + \varphi) \, dt=\frac{eE_0}{m_e\omega} \left(\cos (\omega t + \varphi)-\cos \varphi\right) \tag{3}$$
Taking into account the range of the cosine, the expression in parentheses can take values $[-1-\cos \varphi;\, 1-\cos \varphi]$. One must be careful here: remember that we need the maximum magnitude of the velocity, and that $\cos \varphi$ can be either positive or negative. Finally, analysing (3), we obtain
$$v_{max}=\frac{eE_0}{m_e\omega}(1+|\cos\varphi|).$$
It is clear that the velocity changes periodically with period $T=\frac{2\pi}{\omega}$. Then the average value is
$$\bar{v}=\frac{1}{T}\int_0^T v(t) \, dt \tag{4}$$
I will allow myself to skip the calculation of the integral – remembering that the integral of cosine over its period is zero, this can be done very quickly. Eventually, taking the modulus, we get:
$$\bar{v}=\frac{2eE_0}{m_e\omega}\cos\varphi.$$

There is a typo in the author's answer.