$l = {{\rm{v}}_{\rm{x}}}{\rm{t}} \Rightarrow {\rm{t}} = \frac{l}{{\rm{v}}}$

or $\quad \mathrm{t}=\frac{3 \times 10^{-2} \mathrm{m}}{3 \times 10^{7} \mathrm{m} / \mathrm{s}}=10^{-9} \mathrm{s}$

The force on the electron along $+y$ direction

$\mathrm{F}_{y}=\mathrm{eE}=\frac{\mathrm{eV}}{\mathrm{d}}$

where $\mathrm{V}=550$ volts and $\mathrm{d}=10^{-2} \mathrm{m}$

The acceleration along ty divection is

$a_{y}=\frac{F_{y}}{m}=\frac{e V}{m d}$

For the motion along $+y$ -axis.

${y=0+\frac{1}{2} a_{y} t^{2}}$

or ${\frac{d}{2}=\frac{1}{2}\left(\frac{e V}{m d}\right) t^{2}}$

or $\frac{{\rm{e}}}{{\rm{m}}} = \frac{{{{\rm{d}}^2}}}{{{\rm{V}}{{\rm{t}}^2}}}$

$ = \frac{{{{10}^{ - 4}}}}{{550 \times {{10}^{ - 18}}}} = 1.8 \times {10^{11}}{\rm{C}}/{\rm{kg}}$