Classically permitted position values in a quantum harmonic oscillator
Assume for a minute that instead of having a quantum oscillator with total energy $E$ you had a classical one with the same amount of energy. That energy is the sum of potential and kinetic contributions:
$$E = E_\textrm{p}+E_\textrm{k}.\tag{1}$$
$E_\textrm{k}$ is zero and the total energy equal to the potential energy when the extension (amplitude) is at a maximum, when
$$E = E_{\textrm{p,max}} = \frac12 kx_\mathrm{max}^2\tag{2}$$
which, solving for $x_\mathrm{max},$ leads to
$$x_\mathrm{max} = \sqrt{\frac{2E}{k}}.\tag{3}$$
All that is left to do is to substitute in the right value for $E.$ That is dictated by the expression for the quantized system,
$$E_n = h\nu\left(n + \frac12\right),\tag{4}$$
so that the corresponding classical maximum amplitude is
$$x_\mathrm{max} = \sqrt{\frac{h\nu(2n + 1)}{k}}.\tag{5}$$