Answer
$1.30\;\rm \mu m$
Work Step by Step
We need to find the wavelength of the electrons which is given by
$$\lambda=\dfrac{h}{p}=\dfrac{h}{mv}\tag 1$$
where $v$ is given by $KE=\frac{1}{2}mv^2$, so $v=\sqrt{2KE/m}$
Plugging into (1);
$$\lambda= \dfrac{h}{m\sqrt{2KE/m}} = \dfrac{h}{ \sqrt{2KEm^2/m}}$$
$$\lambda= \dfrac{h}{ \sqrt{2KEm_e }}\tag 2$$
In double slits experiments, for small angles in which $\sin\theta\approx \tan \theta$,
$$d\sin\theta=m\lambda$$
where $\tan\theta=\sin\theta=\dfrac{ x}{L}$
whereas $ x$ is the measured separation distance on the screen and $L$ is the distance between the double slit and the screen.
$$\dfrac{ x d}{L} =m\lambda$$
$$x =\dfrac{ m\lambda L}{d}$$
Thus the separation distance between two peaks is given by
$$\Delta x =\dfrac{ \lambda L}{d}$$
Plugging from (2);
$$\Delta x =\dfrac{ h L}{d\sqrt{2KEm_e }}$$
Plugging the known;
$$\Delta x =\dfrac{ 6.626\times10^{-34}\times 45\times10^{-2}}{2.0\times10^{-6}\sqrt{2\times 45000\times 1.6\times10^{-19}\times 9.11\times10^{-31}}}$$
$$\Delta x =\color{red}{\bf 1.30\times10^{-6}}\;\rm m$$
