Answer
a) $ 2.1\times 10^6\;\rm N/C$
b) $9.37\times 10^7\;\rm m/s$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric field inside a two-plate capacitor is constant and uniform and is given by
$$E=\dfrac{\Delta V}{d}$$
Plug the known;
$$E=\dfrac{25\times 10^3}{1.2\times 10^{-2}}$$
$$E=\color{red}{\bf 2.1\times 10^6}\;\rm N/C$$
$$\color{blue}{\bf [b]}$$
The energy is conserved, so
$$K_i+U_i=K_f+U_f$$
and since its entry speed is close to zero, $K_i=0$
$$0+U_i=K_f+U_f$$
Hence,
$$K_f=-\Delta U=-(-e)\Delta V$$
$$\frac{1}{2}mv_f^2 =e \Delta V$$
$$ v_f =\sqrt{\dfrac{2e \Delta V}{m}}$$
Plug the known;
$$ v_f =\sqrt{\dfrac{2(1.6\times 10^{-19})(25,000)}{9.11\times 10^{-31}}}$$
$$v_f=\color{red}{\bf 9.37\times 10^7}\;\rm m/s$$
