Recall the equation
$\begin{align*}E^2 = m^2 c^4 + p^2 c^2\end{align*}$
where $E$ is the total energy of a particle, $m$ is the rest mass of the particle, $c$ is the speed of light, and $p$ is the particle's relativistic momentum. We are given that $E = 2 mc^2$ , therefore, this equation becomes
$\begin{align*}4 m^2 c^4 = m^2 c^4 + p^2 c^2 \Rightarrow p = \boxed{mc \sqrt{3}}\end{align*}$
Therefore, answer (D) is correct.

Solution to 2001 Problem 32