Answer
a) $1000\;\rm V$
b) $7\;\rm Mm/s$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that
$$ \Delta V_C=Ed$$
Plug the known;
$$ \Delta V_C = (5\times 10^5)(2\times 10^{-3})$$
$$\Delta V_C=\color{red}{\bf 1000}\;\rm V$$
$$\color{blue}{\bf [b]}$$
since the energy is conserved,
$$E_i=E_f$$
$$K_i+U_i=K_f+U_f$$
Assuming the initial potential energy of the electron is zero since it is released from the negative plate
$$K_i+0=K_f+U_f$$
where $U_f=q_eV_C$
Hence,
$$K_i =K_f+qV_C$$
Recalling that $K=\frac{1}{2}mv^2$
$$\frac{1}{2}m_ev_i^2=\frac{1}{2}m_ev_f^2+qV_C$$
$$ v_i^2= v_f^2+\dfrac{2q_eV_C}{m_e}$$
$$ v_i =\sqrt{ v_f^2+\dfrac{2q_eV_C}{m_e}}$$
Plug the known;
$$ v_i =\sqrt{ ( 2.0 \times 10^7 )^2+\dfrac{2 ( 1.6\times10^{-19})(1000)}{(9.11\times10^{-31})}}$$
$$v_i=\color{red}{\bf 7\times 10^6}\;\rm m/s$$
