Answer
${\bf 1.33}\;\rm eV$
Work Step by Step
For an electron in a one-dimensional box of length $ L $, the energy levels are quantized and given by
$$ E_n = \frac{n^2 h^2}{8mL^2} \tag1 $$
where $ n $ is the quantum number (integer), $ h $ is Planck's constant, $ m $ is the mass of the electron, $ L $ is the length of the box.
We also know that
$$E_n=n^2E_1$$
So, the ground-state energy is given by
$$E_1=\dfrac{E_n}{n^2}$$
We can see from the given graph that $n=3$, so
$$E_1=\dfrac{E_3}{3^2}$$
We are given that the electron energy is 12.0 eV, so
$$E_1=\dfrac{12}{9}=\color{red}{\bf 1.33}\;\rm eV$$
