By the kinetic interpretation of temperature, absolute temperature of a given sample of a gas is proportional to the total translational kinetic energy of its molecules.
So, any change in absolute temperature of a gas will contribute to corresponding change in trranslational KE and vice-versa.
As the gas is monoatomic so its degree of freedom will be due to only translational motion, which is three.
So if KE per molecule $=\frac{3}{2} R T$
When, the insulated container stops suddenly its total KE is transferred to gas molecules in the form of translational $\mathrm{KE}$, so increasing in the absolute temperature of gas let it be $\Delta T$ if $n$ is moles of gas.
If $\Delta T=$ change in absolute temperature.
Then, KE of molecules increased due to velocity so
$$
\mathrm{KE}=\frac{1}{2}(m n) v_0^2
$$
where, $n=$ number of moles, $m=$ molar mass of the gas
Increase in translational $(\mathrm{KE})=n \frac{3}{2} R(\Delta T)$
According to kinetic theory Eqs. (i) and (ii) are equal
$$
\begin{aligned}
&\frac{1}{2}(m n) v_0^2=n \frac{3}{2} R(\Delta T) \\
&(m n) v_0^2=n 3 R(\Delta T) \\
&\Delta T=\frac{(m n) v_0^2}{3 n R}=\frac{m v_0^2}{3 R}
\end{aligned}
$$