Answer
$\approx 16^\circ$
Work Step by Step
To find the direction of the final velocity of the electron at the end of the two plates, we need to find its final velocity components at the end of these plates.
And then, we can find the angle by using the tan function since
$$\tan \theta=\dfrac{v_y}{v_x}$$
Hence,
$$\theta=\tan^{-1}\left(\dfrac{v_y}{v_x}\right)\tag 1$$
$\text{Finding $v_x:$}$
We know that the electron accelerated from rest due to a huge potential difference of 2000 V.
This means that its initial kinetic energy was zero and its initial potential energy was maximum.
This means that the whole initial potential energy is converted to kinetic energy in the horizontal direction.
$$PE_i+\overbrace{ KE_i}^{0} =\overbrace{PE_f}^{0} +KE_f$$
Thus,
$$eV_i=\frac{1}{2}m_ev_x^2$$
We know that the horizontal velocity component will not change when the electron is moving between the two plates.
Solving for $v_x$;
$$\dfrac{2eV_i}{m_e }= v_x^2$$
$$v_x=\sqrt{\dfrac{2eV_i}{m_e } }\tag 2$$
$\text{Finding $v_y:$}$
The electric force exerted on the electron due to the potential difference (electric field) is given by
$$F=m_ea_y$$
Then,
$$eE=m_ea_y\tag 3$$
Noting that the vertical acceleration component is given by
$$v_y=\overbrace{ v_{iy}}^{0} +a_yt$$
Initially, the electron has no vertical velocity components, it was moving on a straight line horizontally.
Hence,
$$a_y=\dfrac{v_y}{t}$$
Plugging into (3);
$$eE_y=\dfrac{m_ev_y}{t}\tag 4$$
Now we need to find the time it takes. And since we know that the horizontal component is constant and the distance is about 6.5 cm, so
$$v_x=\dfrac{\Delta x}{t}$$
Thus,
$$t=\dfrac{\Delta x}{v_x}$$
Plugging into (4);
$$eE_y=\dfrac{m_ev_yv_x}{\Delta xt} $$
We also know that the electric field between two plates is given by
$V_{plates}=Ed$ and hence, $E=V_{plates}/d$
Thus,
$$ \dfrac{eV_{plates}}{d}=\dfrac{m_ev_yv_x}{\Delta xt} $$
Solving for $v_y$;
$$v_y=\dfrac{eV_{plates}\Delta x}{ m_ev_xd}\tag 5$$
Plugging (5) into (1);
$$\theta=\tan^{-1}\left(\dfrac{ \dfrac{eV_{plates}\Delta x}{ m_ev_xd}}{v_x}\right)=\tan^{-1}\left(\dfrac{eV_{plates}\Delta x}{v_x^2 m_e d}\right)$$
Plugging $v_x^2$ from (2);
$$\theta =\tan^{-1}\left(\dfrac{eV_{plates}\Delta x}{\dfrac{2eV_i}{m_e } m_e d}\right)=\tan^{-1}\left(\dfrac{ V_{plates}\Delta x}{2 V_i d}\right)$$
Plugging the known;
$$\theta =\tan^{-1}\left(\dfrac{ 250\cdot 6.5\times10^{-2} }{2\cdot 2200\cdot 1.3\times10^{-2}}\right)$$
$$\theta=\color{red}{\bf15.9^\circ }$$
