Answer
${\bf 1.35}\;\rm N/m$
Work Step by Step
We know that
$$ E_n = \left(n - \frac{1}{2}\right)\hbar \omega_e $$
where $n=1,2,3,...$
So,
$$E_{\rm photon}=E_{3}-E_{2}=\dfrac{hc}{\lambda}\tag 1$$
where
$$E_{n+1} = \left(n+1 - \frac{1}{2}\right)\hbar \omega_e= 2.8\;\rm eV $$
$$ \left(n+ \frac{1}{2}\right)\hbar \omega_e= 2.8 \tag 2 $$
and,
$$ E_{n}= \left(n - \frac{1}{2}\right)\hbar \omega_e= 2.0\;\rm eV$$
So,
$$ n =\dfrac{ 2.0}{\hbar \omega_e}+ \frac{1}{2} $$
Plug into (2);
$$ \left(\dfrac{ 2.0}{\hbar \omega_e}+ \frac{1}{2}+ \frac{1}{2}\right)\hbar \omega_e= 2.8 $$
$$ \left(\dfrac{ 2.0}{\hbar \omega_e}+ 1\right)\hbar \omega_e= 2.8 $$
$$ \dfrac{ 2.0\hbar \omega_e}{\hbar \omega_e}+ \hbar \omega_e = 2.8 $$
Hence,
$$ \hbar \omega_e=2.8-2=0.8\;\rm eV$$
Recalling that the angular frequency of a mass on a spring is given by $\omega=\sqrt{k/m}$
$$ \hbar \sqrt{\dfrac{k}{m}} = 0.8 e$$
where $e$ for converting from eV to J.
Solving for $k$;
$$ \sqrt{\dfrac{k}{m}} =\dfrac{ 0.8 e}{\hbar}$$
$$ \dfrac{k}{m} =\dfrac{ 0.8^2 e^2}{\hbar^2}$$
$$ k=\dfrac{ 0.8^2 e^2 m}{\hbar^2}$$
Plug the known;
$$ k=\dfrac{ 0.8^2 (1.6\times 10^{-19})^2 (9.11\times 10^{-31})}{(1.05\times 10^{-34})^2}$$
$$ k=\color{red}{\bf 1.35}\;\rm N/m$$
