Answer
a) $\rm 1.60 \;m$
b) $\rm0.195\;nm$
c) $\rm 0.276 \;nm$
Work Step by Step
We know that the penetration distance is given by
$$ \eta = \frac{\hbar}{\sqrt{2m(U_0 - E)}} $$
We are given that the potential well depth is $ U_0 = 2.00 \; \rm {eV} $, so
$$ \eta = \frac{\hbar}{\sqrt{2m(2.0- E)(1.6\times 10^{-19})}} $$
Plug the other known;
$$ \eta = \frac{(1.055 \times 10^{-34} )}{\sqrt{2(9.11 \times 10^{-31} )(2.0- E)(1.6\times 10^{-19})}} \tag 1$$
$$\color{blue}{\bf [a]}$$
When the electron energy is 0.5 eV,
Plugging into (1);
$$ \eta = \frac{(1.055 \times 10^{-34} )}{\sqrt{2(9.11 \times 10^{-31} )(2.0- 0.5)(1.6\times 10^{-19})}} $$
$$\eta=\color{red}{\bf 1.60 \times 10^{-10} }\;\rm m$$
$$\color{blue}{\bf [b]}$$
When the electron energy is 1.0 eV,
Plugging into (1);
$$ \eta = \frac{(1.055 \times 10^{-34} )}{\sqrt{2(9.11 \times 10^{-31} )(2.0- 1.0)(1.6\times 10^{-19})}} $$
$$\eta=\color{red}{\bf 1.95 \times 10^{-10} }\;\rm m$$
$$\color{blue}{\bf [c]}$$
When the electron energy is 1.5 eV,
Plugging into (1);
$$ \eta = \frac{(1.055 \times 10^{-34} )}{\sqrt{2(9.11 \times 10^{-31} )(2.0- 1.5)(1.6\times 10^{-19})}} $$
$$\eta=\color{red}{\bf 2.76 \times 10^{-10} }\;\rm m$$
