Answer
$\lambda = 68.7~nm$
Work Step by Step
We can find an expression for energy of an electron in this potential well:
$E_n = (\frac{h^2}{8m~L^2})~n^2$
$E_n = \frac{h^2}{8m~L^2}~n^2$
$E_n = \frac{(6.626\times 10^{-34}~J~s)^2}{(8)(9.109\times 10^{-31}~kg)~(250\times 10^{-12}~m)^2}~n^2$
$E_n = (9.640\times 10^{-19}~J)~n^2$
$E_n = (9.640\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})~n^2$
$E_n = (6.02~eV)~n^2$
Note that light with a longer wavelength has less energy than light with a shorter wavelength.
The longest wavelength is absorbed when the electron jumps from the ground state to the state with the smallest energy difference.
We can find the difference in energy between $n=2$ and $n=1$:
$E_2-E_1 = (6.02~eV)(2)^2 - (6.02~eV)(1)^2$
$E_2-E_1 = (6.02~eV)(3)$
$E_2-E_1 = 18.06~eV$
We can find the wavelength associated with this energy:
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(4.136\times 10^{-15}~eV~s)(3.0\times 10^8~m/s)}{18.06~eV}$
$\lambda = 68.7\times 10^{-9}~m$
$\lambda = 68.7~nm$
