In the box $x = 0$ to $x = L$, $V = 0$, and for $x < 0$ and $x > L$, $V = \infty$ (infinite potential well).
The eigenvalues of the Hamiltonian are:

$$E\_n = \frac{n^2 h^2}{8L^2} \, .$$

Since as $V = 0$ in the box, this is kinetic energy only, so:

\begin{align}
\frac{p^2}{2m} &= \frac{n^2 h^2}{8L^2} \\
p^2 &= \frac{n^2 h^2 m}{4L^2} \, .
\end{align}

This is in fact the expectation value of the squared momentum $\langle p^2 \rangle$.
With $\langle p^2 \rangle = m^2\langle v^2 \rangle$ then
$$\langle v^2 \rangle = \frac{n^2 h^2}{4m L^2} \, .$$

We know that for eigenstates of bound particles $\langle v \rangle = 0$, so the average velocity of the particle in the well is $0$, which makes perfect sense.
So what is the physical meaning of
$$\sqrt{\langle v^2 \rangle} = \sqrt{ \frac{n^2 h^2}{4m L^2}} = \frac{1}{\sqrt{m}} \frac{nh}{2L} \, ?$$
Or am I thinking too "classical"?