Answer
$\approx {\bf 9\times 10^{-8} }\;\rm eV$
Work Step by Step
We need to find the ground-state energy of the sodium ion confined in this box.
We know that the energy levels of a particle in a box are given by
$$E_n = \dfrac{n^2 h^2}{8mL^2}$$
Where: $ E_n $ is the energy of the $n$-th energy level and since the author told us to find that of the ground state, $n=1$, $h$ is Planck's constant, $m$ is the mass of the particle which here is the mass of the sodium ion, and $L$ is the length of the box.
Plug the known;
$$E_1 = \dfrac{ h^2}{8mL^2}$$
$$E_1 = \dfrac{ (1)^2 (6.626 \times 10^{-34})^2}{8m_{\rm Na}(10\times 10^{-9})^2}$$
where the mass of a sodium ion is $ \approx 3.82 \times 10^{-26} \; \rm{kg}$
$$E_1 = \dfrac{ (1)^2 (6.626 \times 10^{-34})^2}{8 (3.82 \times 10^{-26} )(10\times 10^{-9})^2}$$
$$E_1=\bf 1.437\times 10^{-26}\;\rm J\approx \color{red}{\bf 9\times 10^{-8} }\;\rm eV$$
