Answer
${\bf 519}\;\rm nm$
Work Step by Step
We know that
$$ E_n = \left(n - \frac{1}{2}\right)\hbar \omega_e $$
where $n=1,2,3,...$
So, the longest wavelength of light that can be absorbed is given by
$$E_{\rm photon}=E_{n+1}-E_{n}=\dfrac{hc}{\lambda}\tag 1$$
where
$$E_{n+1}-E_{n}= \left(n+1 - \frac{1}{2}\right)\hbar \omega_e- \left(n - \frac{1}{2}\right)\hbar \omega_e =\hbar \omega_e $$
Plug into (1);
$$\hbar \omega_e =\dfrac{hc}{\lambda}$$
Solving for $\lambda$;
$$ \lambda=\dfrac{hc}{\hbar \omega_e}=\dfrac{2\pi hc}{h\omega_e}=\dfrac{2\pi c}{ \omega_e}$$
Recalling that the angular frequency of a mass on a spring is given by $\omega=\sqrt{k/m}$.
So,
$$ \lambda= \left[2\pi c \right]\cdot \sqrt{\dfrac{m}{k}}$$
Plug the known;
$$ \lambda= \left[2\pi (3\times 10^8) \right]\cdot \sqrt{\dfrac{(9.11\times 10^{-31})}{12}}$$
$$ \lambda =\bf 5.19 \times 10^{-7} \;\rm m=\color{red}{\bf 519}\;\rm nm$$
