Answer
$3.218\;eV$
Work Step by Step
The quantized energies for an electron trapped in a three-dimensional infinite potential well that forms a e rectangular box are
$E_{nx,ny,nz}=\frac{h^2}{8m}\Big(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}+\frac{n_z^2}{L_z^2}\Big),$
where $m$ is the electron mass and $n_x$ is a quantum number for well width $L_x$, $n_y$ is a quantum number for well width $L_y$ and $n_z$ is a quantum number for well width $L_z$.
The ground state is the $(1,1)$ state, with an energy of
$E_{1,1,1}=\frac{h^2}{8m}\Big(\frac{1}{L_x^2}+\frac{1}{L_y^2}+\frac{1}{L_z^2}\Big)$
Substituting, $L_x=800\;pm=800\times 10^{-12}\; m$,
$L_y=1600\;pm=1600\times 10^{-12}\; m$,
and $L_z=390\;pm=390\times 10^{-12}\; m$ we get the ground state energy of the electron in the given rectangular box
$E_{1,1,1}=\frac{(6.63\times 10^{-34})^2}{8\times 9.1\times 10^{-31}}\Big\{\frac{1}{(800\times 10^{-12})^2}+\frac{1}{(1600\times 10^{-12})^2}+\frac{1}{(390\times 10^{-12})^2}\Big\}$
$E_{1,1,1}\approx 5.149\times 10^{-19}\;J$
$E_{1,1,1}\approx 3.218\;eV$
