Given, the half-life of neutron, $t_{1 / 2}=693 \mathrm{sec}$ Kinetic energy of neutron $=0.084 \mathrm{eV}$
$\begin{aligned}
\frac{1}{2} m v^2 & =0.084 \times 1.6 \times 10^{-19} \\
v^2 & =\frac{2 \times 0.084 \times 1.6 \times 10^{-19}}{m} \\
& =\frac{2 \times 0.084 \times 1.6 \times 10^{-19}}{1.68 \times 10^{-27}}=0.16 \times 10^8 \\
v & =0.4 \times 10^4 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Time taken by neutron to travelled in $1 \mathrm{~km}$,
$t=\frac{1000}{0.4 \times 10^4} \Rightarrow t=0.25 \mathrm{~s}$
By Radioactive Decay's law,
$\begin{aligned}
N & =N_0\left(\frac{1}{2}\right)^n \Rightarrow \frac{N}{N_0}=\left(\frac{1}{2}\right)^n \\
\frac{N_0}{N} & =2^n
\end{aligned}$
Taking log on the both sides, we get
$\begin{aligned}
\ln \frac{N_0}{N}=\ln 2^n & =n \ln 2=\frac{t}{t_{1 / 2}} \ln 2=\frac{0.25 \times 0.693}{693} \\
& =0.25 \times 10^{-5} \quad[\because \ln 2=0.693]
\end{aligned}$