Answer
At a wavelength of $~~547~nm~~$ visible light that is reflected from the two surfaces will undergo fully constructive interference.
Work Step by Step
Since the index of refraction of the film has a greater index of refraction than air, there is a phase change of $\frac{\lambda}{2}$ due to reflection with the front surface of the film.
Since the index of refraction of the film has a greater index of refraction than the air, there is no phase change due to reflection with the back surface of the film.
To undergo fully constructive interference, the path length difference should be $(m+0.5)~\lambda$, where $m$ is an integer.
We can find $\lambda$:
$2L = (m+0.5)~\lambda$, where $m$ is an integer
$\lambda = \frac{2L}{m+0.5}$
$\lambda = \frac{(2)(410~nm)}{m+0.5}$
$\lambda = \frac{820~nm}{m+0.5}$
$\lambda = 1640~nm, 547~nm, 328~nm,...$
Note that light in the visible range has a wavelength of approximately $400~nm$ to $700~nm$
At a wavelength of $~~547~nm~~$ visible light that is reflected from the two surfaces will undergo fully constructive interference.
