Answer
${\bf 2.25}\;\rm N/m$
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
$$E_{\rm photon}=E_{3}-E_{2}=\dfrac{hc}{\lambda}\tag 1$$
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);
$$\hbar \omega_e =\dfrac{hc}{\lambda}$$
Recalling that the angular frequency of a mass on a spring is given by $\omega=\sqrt{k/m}$.
So,
$$\hbar \sqrt{\dfrac{k}{m}} =\dfrac{hc}{\lambda}$$
Solving for $k$;
$$ \sqrt{\dfrac{k}{m}} =\dfrac{hc}{\hbar\lambda}=\dfrac{2\pi hc}{h\lambda}= \dfrac{ 2\pi c}{\lambda}$$
$$ \dfrac{k}{m} = \dfrac{ 4\pi^2 c^2}{\lambda^2}$$
$$k = \dfrac{ 4\pi^2m c^2}{\lambda^2}$$
Plug the known;
$$k = \dfrac{ 4 \pi^2 (9.11\times 10^{-31})(3\times 10^8)^2}{(1200\times 10^{-9})^2}$$
$$ k =\color{red}{\bf 2.25}\;\rm N/m$$
