Answer
$2.3\,pm$
Work Step by Step
A helium atom with energy $1.0 \,eV$ below $U_{0}$ has $U_{0}-E=1.0\,eV$.
The penetration distance $\eta=\frac{\hbar}{\sqrt {2m(U_{0}-E)}}$
$=\frac{1.05\times10^{-34}\,J\cdot s}{\sqrt {2(4\times1.66\times10^{-27}\,kg)(1.0\,eV\times\frac{1.6\times10^{-19}\,J}{1.0\,eV})}}$
$=2.3\times10^{-12}\,m=2.3\,pm$
