Answer
$ 2.85\times 10^5 \;\rm N/C$
Work Step by Step
First, we need to recall Newton's second law to find the force exerted on the electron.
$$F=m_ea$$
and since it travels in a uniform electric field, the force is then given by $F=q_eE$.
$$q_eE=m_ea$$
Solving for $E$,
$$ E=\dfrac{m_ea}{q_e}\tag 1$$
where $a$ is found by the kinematic formula of $v_f^2=v_i^2+2ad$, so
$$a=\dfrac{v_f^2-v_i^2}{2d}$$
Plugging into (1),
$$ E=\dfrac{m_e }{q_e} \dfrac{v_f^2-v_i^2}{2d}$$
Plug the known;
$$ E=\dfrac{9.11\times 10^{-31}}{1.6\times 10^{-19}}\cdot \dfrac{(4\times 10^7)^2-(2\times 10^7)^2}{2(1.2\times 10^{-2})}$$
$$E=\color{red}{\bf 2.85\times 10^5}\;\rm N/C$$
