Answer
${\bf 3400}\;\rm eV$
Work Step by Step
To solve this problem, we can use the principle of conservation of energy. Initially, the proton has only kinetic energy, and as it moves away from the charged glass bead, some of this kinetic energy is converted into electric potential energy due to the work done by the electric force between the proton and the charged bead.
$$U_i+K_i=U_f+K_f$$
The bead here works as the Earth and the proton here works as baseball that is fired upward. This means that the glass bead energy and charge are constants.
$$\dfrac{kQq}{r_i}+K_i=\dfrac{kQq}{r_f}+K_f$$
where $Q$ is the charge of the bead, $q$ is the charge of the proton, $r_i=r_{\rm bead}$, and $r_f=r_{\rm bead}+2\;\rm mm$
So, the final kinetic energy of the proton at $r_f$ is
$$K_f=\dfrac{kQq}{r_{\rm bead}}+K_i-\dfrac{kQq}{r_{\rm bead}+0.002}$$
$$K_f=K_i+kQq\left[\dfrac{1}{r_{\rm bead}}-\dfrac{1}{r_{\rm bead}+0.002}\right]$$
Plug the known; and remember to unify the unit of the energy.
$$K_f=(520)+(9\times 10^9)(0.2\times 10^{-9})(1.6\times 10^{-19})\left[\dfrac{1}{(0.5\times 10^{-3})}-\dfrac{1}{(0.5\times 10^{-3})+0.002}\right]\dfrac{1}{1.6\times 10^{-19}}$$
$$K_f=\color{red}{\bf 3400}\;\rm eV$$
