Answer
$\lambda = 608~nm$
Work Step by Step
Note that light in the visible range has a wavelength of approximately $400~nm$ to $700~nm$
Since $n_2 \gt n_1$, there is a phase shift of $\frac{1}{2}~\lambda$ of ray $r_1$ due to reflection.
Since $n_3 \lt n_2$, there is no phase shift of ray $r_2$ 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 $\lambda$:
$2L = \frac{(m+\frac{1}{2})~\lambda}{n_2},$ where $m$ is an integer
$\lambda = \frac{2~L~n_2}{(m+\frac{1}{2})}$
$\lambda = \frac{(2)~(285~nm)~(1.60)}{(m+\frac{1}{2})}$
$\lambda = \frac{(912~nm)}{(m+\frac{1}{2})}$
$\lambda = 1824~nm, 608~nm, ...$
Since $608~nm$ is in the visible range, $~~\lambda = 608~nm$
