Answer
The film thickness is $~~338~nm$
Work Step by Step
Note that there is a phase shift of $\frac{1}{2}~\lambda$ of the ray striking the front of the film due to reflection.
If there is a maximum, the path length difference must be $~~(m+\frac{1}{2})~\frac{\lambda}{n_2}~~$, where $m$ is an integer.
If there is a minimum, the path length difference must be $~~(m)~\frac{\lambda}{n_2}~~$, where $m$ is an integer.
We can find possible values for the thickness $L$:
$2L = \frac{(m+\frac{1}{2})~\lambda}{n_2},$ where $m$ is an integer
$L = \frac{(m+\frac{1}{2})~\lambda}{2~n_2}$
$L = \frac{(m+\frac{1}{2})~(600~nm)}{(2)~(1.33)}$
$L = (m+\frac{1}{2})~(225.56~nm)$
$L = 113~nm, 338~nm, 564~nm,...$
We can find possible values for the thickness $L$:
$2L = \frac{(m)~\lambda}{n_2},$ where $m$ is an integer
$L = \frac{(m)~\lambda}{2~n_2}$
$L = \frac{(m)~(450~nm)}{(2)~(1.33)}$
$L = (m)~(169.17~nm)$
$L = 169~nm, 338~nm, 508~nm, ...$
We can see that the film thickness must be $~~338~nm$
