Answer
a) ${\bf 1.33\times 10^8}\;\rm m/s$
b) ${\bf 5.47}\;\rm \mu m $
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the kinetic energy is given by
$$K=\frac{1}{2}mv^2$$
So the speed of the electron is given by
$$v=\sqrt{\dfrac{2K}{m}}$$
Plug the known;
$$v=\sqrt{\dfrac{2(50\times 10^{3}\times 1.6\times 10^{-19})}{(9.11\times 10^{-31})}}$$
$$v=\color{red}{\bf 1.33\times 10^8}\;\rm m/s$$
$$\color{blue}{\bf [b]}$$
Recalling that the spacing between fringes in a double-slit interference experiment is given by
$$\Delta y=\dfrac{\lambda L}{d}\tag 1$$
where $L$ is the distance between the double-slit and the screen, $d$ is the slit separation, and $\lambda$ is the wavelength of the electrons which we can find by using de Broglie wavelength's formula.
$$\lambda=\dfrac{h}{p}=\dfrac{h}{mv}$$
Plug into (1);
$$\Delta y=\dfrac{h L}{mvd} $$
Plug the known;
$$\Delta y=\dfrac{(6.63\times 10^{-34})(1)}{(9.11\times 10^{-31})(1.33\times 10^8)(1.0\times 10^{-6}) }$$
$$\Delta y=\bf 5.47\times 10^{-6}\;\rm m=\color{red}{\bf 5.47}\;\rm \mu m $$
