Answer
${\bf 0.739}\;\rm nm$
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.
Now we need to find the energy difference corresponding to the absorption which is the energy of the absorbed photon.
Assuming that the transition is from $n=1$ to $n=2$ since the absorption occurs from the ground level $n=1$.
$$ E_{\rm photon}=\Delta E_{\rm rlrctron} = E_2 - E_1$$
$$ \frac{hc}{\lambda} = \frac{2^2h^2}{8mL^2} - \frac{1^2\cdot h^2}{8mL^2} = \frac{3h^2}{8mL^2}$$
Solving for $L$;
$$ L^2 = \frac{3h\lambda}{8mc} $$
$$ L =\sqrt{ \frac{3h\lambda}{8mc} }$$
Substitute the known:
$$
L = \sqrt{\frac{3 \times (6.63 \times 10^{-34}) \times (600 \times 10^{-9})}{8 (9.11 \times 10^{-31}) (3.00 \times 10^8)} }$$
$$L=\bf 7.388\times10^{-10}\;\rm m=\color{red}{\bf 0.739}\;\rm nm$$
