Answer
$L = 329~nm$
Work Step by Step
Note that light in the visible range has a wavelength of approximately $400~nm$ to $700~nm$
Since $n_2 \lt n_1$, there is no phase shift of ray $r_1$ due to reflection.
Since $n_3 \gt n_2$, there is a phase shift of $\frac{1}{2}~\lambda$ 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 thickness $L$:
$2L = \frac{(m+\frac{1}{2})~\lambda}{n_2},$ where $m$ is an integer
$L = \frac{(m+\frac{1}{2})~\lambda}{2~n_2}$
$L = \frac{(m+\frac{1}{2})~(587~nm)}{(2)~(1.34)}$
$L = (m+\frac{1}{2})~(219.03~nm)$
$L = 110~nm, 329~nm,...$
Since we are looking for the second least thickness, $~~L = 329~nm$
