Answer
$1.2 \times 10^7\;\rm m/s$
Work Step by Step
Since the electric field is uniform inside the capacitor, the force exerted on the electron when it is moving between the plates is constant as well.
This means that the acceleration of the electron is constant.
According to Newton's second law, the acceleration of the electron is given by
$$\sum F_y=q_e E=m_ea_y$$
Thus,
$$a_y =\dfrac{q_e E}{m_e}\tag 1$$
The minimum initial speed $v_0$ the electron can have without hitting the negative plate means the maximum height of the electron is 2 cm (which is the distance between the two plates) from the postive plate.
The maximum height of $h=0.02$ m occurs when $v_{fy}=0$ and the electron starts to fall back again toward the positive plate.
$$v_{fy}^2=v_{iy}^2-2a_y h$$
where $h$ is the distance between the two plates or the maximum height of the electron, and $v_{iy}=v_0\sin45^\circ$
$$0^2=v_0^2\sin^245^\circ-2a_y h$$
Hence,
$$ v_0 =\sqrt{\dfrac{2a_y h}{\sin^245^\circ}}$$
Plug from (1),
$$ v_0 =\sqrt{\dfrac{2 q_ehE}{\sin^245^\circ m_e}}$$
Plug the known;
$$ v_0 =\sqrt{\dfrac{2 (1.6\times 10^{-19})(0.02)(1\times 10^4)}{\sin^245^\circ (9.11\times 10^{-31})}}$$
$$v_0=\color{red}{\bf 1.2 \times 10^7}\;\rm m/s$$
