Answer
$\bf 0.0382\;\rm eV$
Work Step by Step
We know that the penetration distance is given by
$$ \eta = \frac{\hbar}{\sqrt{2m(U_0 - E)}} $$
So the electron energy is given by
$$ \eta^2 = \frac{\hbar^2}{2m(U_0 - E)} $$
$$ U_0 - E= \frac{\hbar^2}{2m\;\eta^2 } $$
Plug the known
$$ U_0 - E= \frac{(1.055 \times 10^{-34})^2}{2(9.11 \times 10^{-31})(1\times 10^{-9})^2 } =\bf 6.11 \times 10^{-21}\;\rm J$$
$$ U_0 - E =\color{red}{\bf 0.0382} \;\rm eV$$
Therefore, the electron is $\bf 0.0382$ eV below the $U_0$.
