The wave-function amplitude decays as $e^{-\kappa x}$, so its e-folding penetration depth is
$$
\boxed{\Delta x=\kappa^{-1}=\frac\hbar{\sqrt{2m(V_0-E)}}.}
$$
The <probability density> decays as $e^{-2\kappa x}$ and has the smaller e-folding depth $1/(2\kappa)$. Specifying which quantity decays avoids a factor-of-two ambiguity in the term penetration depth.

Solution

Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S3 / 1 / c / Solution

 Solution (source code)