Answer
The thickness of the acetone film is $~~840~nm$
Work Step by Step
There is a phase shift of $\frac{1}{2}~\lambda$ due to reflection at the front surface of the film.
There is a phase shift of $\frac{1}{2}~\lambda$ due to reflection at the back surface of the film.
We can think of the net phase shift due to reflection as zero.
If there is a maximum, the path length difference must be $~~(m)~\frac{\lambda}{1.25}~~$, where $m$ is an integer.
If there is a minimum, the path length difference must be $~~(m+0.5)~\frac{\lambda}{1.25}~~$, where $m$ is an integer.
We can find possible values for the thickness $L$ of the film using the maximum:
$2L = \frac{(m)~\lambda}{1.25},$ where $m$ is an integer
$L = \frac{(m)~\lambda}{2~(1.25)}$
$L = \frac{(m)~(700~nm)}{(2)~(1.25)}$
$L = (m)~(280~nm)$
$L = 280~nm, 560~nm, 840~nm,...$
We can find possible values for the thickness $L$ of the film using the minimum:
$2L = \frac{(m+0.5)~\lambda}{1.25},$ where $m$ is an integer
$L = \frac{(m+0.5)~\lambda}{2~(1.25)}$
$L = \frac{(m+0.5)~(600~nm)}{(2)~(1.25)}$
$L = (m+0.5)~(240~nm)$
$L = 120~nm, 360~nm, 600~nm, 840~nm,...$
We can see that the thickness of the acetone film must be $~~840~nm$
