Answer
a) ${\bf 4.95}\;\rm eV $
b) ${\bf 4.80 }\;\rm eV $
c) ${\bf 4.55 }\;\rm eV $
Work Step by Step
We know that the probability that an electron will tunnel is given by
$$P_{\text{tunnel}} = e^{-2w/\eta} $$
where $\eta = \dfrac{\hbar}{\sqrt{2m(U_0 - E)}} $ , so
$$P_{\text{tunnel}} = e^\left[\dfrac{-2w\sqrt{2m(U_0 - E)}}{\hbar} \right] $$
Taking the natural logarithm.
$$\ln(P_{\text{tunnel}}) = \dfrac{-2w\sqrt{2m(U_0 - E)}}{\hbar} $$
Solving for $E$ since we need to find the electron energy.
$$\dfrac{\hbar\; \ln(P_{\text{tunnel}})}{-2w} = \sqrt{2m(U_0 - E)}$$
Squaring both sides
$$\dfrac{\hbar^2\; \ln(P_{\text{tunnel}})^2}{4w^2} = 2m(U_0 - E) $$
$$U_0 - E=\dfrac{\hbar^2\; \ln(P_{\text{tunnel}})^2}{8mw^2} $$
Thus,
$$ E=U_0 -\dfrac{\hbar^2\; \ln(P_{\text{tunnel}})^2}{8mw^2} $$
Plug the known;
$$\boxed{ E=(5\times 1.6\times 10^{-19})-\dfrac{(1.05\times 10^{-34})^2\; \ln(P_{\text{tunnel}})^2}{8(9.11\times 10^{-31})(1\times 10^{-9})^2}} $$
$$\color{blue}{\bf [a]}$$
when $P_{\text{tunnel}}=10\%=0.1$, plug into the boxed formula,
$$ E=(5\times 1.6\times 10^{-19})-\dfrac{(1.05\times 10^{-34})^2\; \ln(0.1)^2}{8(9.11\times 10^{-31})(1\times 10^{-9})^2} $$
$$E=\bf 7.92 \times 10^{-19}\;\rm J=\color{red}{\bf 4.95}\;\rm eV $$
$$\color{blue}{\bf [b]}$$
when $P_{\text{tunnel}}=1\%=0.01$, plug into the boxed formula,
$$ E=(5\times 1.6\times 10^{-19})-\dfrac{(1.05\times 10^{-34})^2\; \ln(0.01)^2}{8(9.11\times 10^{-31})(1\times 10^{-9})^2} $$
$$E=\bf 7.68 \times 10^{-19}\;\rm J=\color{red}{\bf 4.80 }\;\rm eV $$
$$\color{blue}{\bf [c]}$$
when $P_{\text{tunnel}}=0.1\%=0.001$, plug into the boxed formula,
$$ E=(5\times 1.6\times 10^{-19})-\dfrac{(1.05\times 10^{-34})^2\; \ln(0.001)^2}{8(9.11\times 10^{-31})(1\times 10^{-9})^2} $$
$$E=\bf 7.28 \times 10^{-19}\;\rm J=\color{red}{\bf 4.55 }\;\rm eV $$
