Answer
$\lambda = 560~nm$
Work Step by Step
Note that light in the visible range has a wavelength of approximately $400~nm$ to $700~nm$
Since $n_3 \gt n_2$ and $n_1 \gt n_2$, there are two phase shifts of $\frac{1}{2}~\lambda$ each due to reflection.
Note that we can consider the net phase shift to be zero.
Since we are looking for a maximum, the path length difference must be $~~(m)~\frac{\lambda}{n_2}~~$, where $m$ is an integer.
We can find the required wavelength:
$2L = \frac{(m)~\lambda}{n_2},$ where $m$ is an integer
$\lambda = \frac{2~L~n_2}{m}$
$\lambda = \frac{(2)~(200~nm)~(1.40)}{m}$
$\lambda = \frac{560~nm}{m}$
$\lambda = 560~nm,...$
Since $560~nm$ is in the visible range, $~~\lambda = 560~nm$
