\(\displaystyle{a}={\frac{{{q}{E}}}{{{m}}}}={\frac{{{1.6}\cdot{10}^{{-{19}}}{5}\cdot{10}^{{3}}}}{{{9.1}\cdot{10}^{{-{31}}}}}}\)

\(\displaystyle{a}={\frac{{{q}{E}}}{{{m}}}}={8.79}\cdot{10}^{{{15}}}\frac{{m}}{{s}^{{2}}}\)

We can find the exit velocity:

\(\displaystyle{v}^{{2}}={{v}_{{0}}^{{2}}}+{2}{a}{d}\)

\(\displaystyle{v}^{{2}}={\left({1}\cdot{10}^{{7}}\right)}^{{2}}+{2}\cdot{8.79}\cdot{10}^{{{15}}}\cdot{0.0049}\)

\(\displaystyle{v}={1.364}\cdot{10}^{{7}}\frac{{m}}{{s}}\)

\(\displaystyle\theta={{\tan}^{{-{1}}}{\left({\frac{{{v}}}{{{v}_{{0}}}}}\right)}}\)

\(\displaystyle\theta={{\tan}^{{-{1}}}{\left({\frac{{{1.364}\cdot{10}^{{7}}}}{{{1}\cdot{10}^{{7}}}}}\right)}}\)

\(\displaystyle\theta={53.75}^{\circ}\)