Answer
$E_1 = 1.9~GeV$
Work Step by Step
Note that in the ground-state, $n=1$
We can find the energy of an electron in the ground-state:
$E_n = (\frac{h^2}{8m~L^2})~n^2$
$E_1 = \frac{h^2}{8m~L^2} (1)^2$
$E_1 = \frac{(6.626\times 10^{-34}~J~s)^2}{(8)(9.109\times 10^{-31}~kg)~(1.4\times 10^{-14}~m)^2}$
$E_1 = (3.0739\times 10^{-10}~J)$
$E_1 = (3.0739\times 10^{-10}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E_1 = 1.9\times 10^9~eV$
$E_1 = 1.9~GeV$
