Answer
$\lambda = 409~nm$
Work Step by Step
Note that light in the visible range has a wavelength of approximately $400~nm$ to $700~nm$
Since $n_3 \gt n_2$, there is a phase shift of $\frac{1}{2}~\lambda$ due to reflection.
Since we are looking for a maximum, the path length difference must be $~~(m+\frac{1}{2})~\frac{\lambda}{n_2}~~$, where $m$ is an integer.
We can find the required wavelength:
$2L = \frac{(m+\frac{1}{2})~\lambda}{n_2},$ where $m$ is an integer
$\lambda = \frac{2~L~n_2}{m+\frac{1}{2}}$
$\lambda = \frac{(2)~(210~nm)~(1.46)}{m+\frac{1}{2}}$
$\lambda = \frac{613.2~nm}{m+\frac{1}{2}}$
$\lambda = 1226.4~nm, 409~nm,...$
Since $409~nm$ is in the visible range, $~~\lambda = 409~nm$
