Answer
$72.2~eV~~$ of energy must be transferred to the electron.
Work Step by Step
We can find an expression for energy of an electron in this potential well:
$E_n = (\frac{h^2}{8m~L^2})~n^2$
$E_n = \frac{h^2}{8m~L^2}~n^2$
$E_n = \frac{(6.626\times 10^{-34}~J~s)^2}{(8)(9.109\times 10^{-31}~kg)~(250\times 10^{-12}~m)^2}~n^2$
$E_n = (9.640\times 10^{-19}~J)~n^2$
$E_n = (9.640\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})~n^2$
$E_n = (6.02~eV)~n^2$
We can find the energy in the first excited state where $n=2$:
$E_2 = (6.02~eV)~(2)^2 = 24.1~eV$
We can find the energy in the third excited state where $n=4$:
$E_4 = (6.02~eV)~(4)^2 = 96.3~eV$
We can find the difference in energy:
$E_4-E_2 = (96.3~eV) - (24.1~eV) = 72.2~eV$
$72.2~eV~~$ of energy must be transferred to the electron.
