Answer
See the detailed answer below.
Work Step by Step
We are dealing with an electron in a finite potential well.
$$\color{blue}{\bf [a]}$$
The wave function $ \psi(x) $ for the electron in the classically forbidden region at $ x = L $ is given by
$$
\psi(x) = \psi_{\text{edge}} \;e^{-(x - L)/\eta}\tag 1
$$
The ratio of the probability densities at two positions, $ x = L $ and $ x = L + d $, is:
$$
\frac{P(d + L)}{P(L)} = \frac{|\psi(d + L)|^2}{|\psi(L)|^2}=10\%=0.10
$$
Substitute from (1);
$$
\frac{P(d + L)}{P(L)} = \frac{|\psi_{\text{edge}} \; e^{-d/\eta}|^2}{|\psi_{\text{edge}}|^2\;e^0} = e^{-2d/\eta}=0.10
$$
$$ e^{-2d/\eta}=0.10\tag 2$$
Now we need to find $\eta$ which is given by
$$
\eta = \frac{\hbar}{\sqrt{2m(U_0 - E)}}
$$
where $ U_0 - E = 2.7 \; \text{eV} $
The electron in sodium metal has an energy of $\rm -2.7 eV$, which means it is trapped inside the metal. The energy difference, $ U_0 - E = 2.7 \, \text{eV} $, where $ U_0 $ is the surface's potential energy barrier, shows how much energy is keeping the electron inside. So the electron's penetration depth is
$$
\eta = \frac{(1.05 \times 10^{-34})}{\sqrt{2 (9.11 \times 10^{-31}) (2.7\times 1.6\times 10^{-19})}}=\bf 1.18 \times 10^{-10} \;\rm m
$$
$$
\eta =\bf \; 0.118 \; \rm \text{nm}\tag 3
$$
Solving (2) for $ d $ by taking the natural logarithm of both sides;
$$
\dfrac{-2d}{\eta }= \ln(0.10)
$$
$$d=\dfrac{ \eta\;\ln(0.10)}{-2}$$
Substitute from (3);
$$d=\dfrac{ (0.118 )\;\ln(0.10)}{-2}=\color{red}{\bf 0.136}\;\rm nm$$
$$\color{blue}{\bf [b]}$$
A typical atomic diameter is approximately $ 2a_B $, where $ a_B $ is the Bohr radius, which is about $ 0.1 \; \text{nm} $.
The penetration distance of $ 0.118 \; \text{nm} $ is on the same order of magnitude as the atomic diameter.
Thus, the penetration distance of the electron beyond the surface is roughly one atomic diameter.
