Answer
$$\frac{3}{2}\left(8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}\right)(300 \mathrm{K}) \approx 0.04 \mathrm{\ eV}
$$
Work Step by Step
At $T=300 \mathrm{K}, \text { the average kinetic energy of the neutrons is (using Eq. } 20-24)$:
$$
K_{\mathrm{apg}}=\frac{3}{2} K T=\frac{3}{2}\left(8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}\right)(300 \mathrm{K}) \approx 0.04 \mathrm{eV}
$$
