(a)
Hint: \(K=\frac{p^{2}}{2m}\)
Step 1: Find the momentum of a neutron.
According to De Broglie
           \(\lambda=\frac{h}{p}\\p=\frac{h}{\lambda}\)
Step 2: Find the kinetic energy of a neutron.
\(K=\frac{p^2}{2m}=\frac{h^2}{2m\lambda^2}\) 
$\begin{array}{l}=\frac{(6.63\times {10}^{-34}{)}^2}{2\times (1.40\times {10}^{-10}{)}^2\times 1.66\times {10}^{-27}}=6.75\times {10}^{-21}J\\ =\frac{6.75\times {10}^{-21}}{1.6\times {10}^{-19}}=4.219\times {10}^{-2}eV\end{array}$
Hence, the kinetic energy of the neutron is 6.75 x 10^-21 J or 4.219 x 10^-2 eV.

(b)
Hint: \(\lambda^{\prime}=\frac{h}{\sqrt{2 K m }}\)

Step 1: Find the average kinetic energy of the neutron.
The average kinetic energy of the neutron:
$\begin{array}{l}{K}^{\mathrm{}}=\frac{3}{2}kT\\ =\frac{3}{2}\times 1.38\times {10}^{-23}\times 300=6.21\times {10}^{-21}J\end{array}$

Step 2: Find the wavelength of a neutron.
The relation for the de Broglie wavelength is given as ${\lambda }^{\mathrm{\prime }}=\frac{h}{\sqrt{2K^{\mathrm{}}{m}_{}}}$
$\begin{array}{l}{\lambda }^{\mathrm{\prime }}=\frac{6.63\times {10}^{-34}}{\sqrt{2\times 6.21\times {10}^{-21}\times 1.66\times {10}^{-27}}}\\ =1.46\times {10}^{-10}m=0.146nm\end{array}$
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.