Answer
$6.21\times 10^{-21}\;J$
Work Step by Step
Neutrons in thermal equilibrium with matter have an average kinetic energy $E_K=\frac{3}{2}kT$, where $k$ is the Boltzmann constant and $T$ is the temperature of the environment of the neutrons. $T$ may be taken to be $300\;K$.
$\therefore$ The average kinetic energy of such a neutron is
$E_K=\frac{3}{2}\times1.38\times10^{-23}\times300\;J$
$E_K=6.21\times 10^{-21}\;J$
