Answer
At the wavelength $~~~450~nm~~~$ the reflected light undergoes fully destructive interference.
Work Step by Step
We can assume that the index of refraction of the film is greater than the index of refraction of air.
There is a phase shift of $\frac{1}{2}~\lambda$ due to reflection at the front surface of the film.
There is no phase shift due to reflection at the back surface of the film.
If there is a maximum (fully constructive interference), the path length difference must be $~~(m+0.5)~\frac{\lambda}{n_2}~~$, where $m$ is an integer.
We can find the index of refraction of the film:
$2L = \frac{(m+0.5)~\lambda}{n_2},$ where $m$ is an integer
$n_2 = \frac{(m+0.5)~\lambda}{2~L}$
$L = \frac{(m+0.5)~(600.0~nm)}{(2)~(272.7~nm)}$
$L = (m+0.5)~(1.1)$
$L = 0.55, 1.65, 2.75,...$
The most reasonable value for the index of refraction of the film is $1.65$
If there is a minimum (fully destructive interference), the path length difference must be $~~(m)~\frac{\lambda}{n_2}~~$, where $m$ is an integer.
We can find the wavelength $\lambda$ where there is fully destructive interference:
$2L = \frac{(m)~\lambda}{1.65},$ where $m$ is an integer
$\lambda = \frac{2L~(1.65)}{(m)}$
$\lambda = \frac{(2)(272.7~nm)~(1.65)}{(m)}$
$\lambda = \frac{900~nm}{(m)}$
$\lambda = 900~nm, 450~nm, 300~nm,...$
Note that light in the visible range has a wavelength of approximately $400~nm$ to $700~nm$
Since $450~nm$ is in the visible range, $~~\lambda = 450~nm$
At the wavelength $~~~450~nm~~~$ the reflected light undergoes fully destructive interference.
