Answer
$1.35$
Work Step by Step
As we see in the figures below, we have two cases here.
In the first case, when the index of refraction of the film is less than that of the glass but higher than that of air, so we got one reflected ray with a $\pi$ phase change [which means that the rays are reflected out of phase].
The path length difference for this case when we need a destructive interference is given by
$$2t= m \lambda_n=\dfrac{m\lambda}{n}$$
Solving for $n$;
$$n=\dfrac{m\lambda}{2t}$$
Plugging the given; and for minimum $n$, we use $m=1$
$$n=\dfrac{1\times 675\times 10^{-9}}{2\times 125\times 10^{-9}}=\color{red}{\bf 2.7}$$
which is a very huge index of refraction and it seems that it is unlikely to be used there.
---
In the second case, when the index of refraction of the film is greater than that of the glass, so we got two reflected rays with $\pi$ phase changes [which means that both rays are reflected in phase].
The path length difference for this case when we need a destructive interference is given by
$$2t=\left(m+\frac{1}{2}\right)\lambda_n=\dfrac{\left(m+\frac{1}{2}\right)\lambda}{n}$$
Solving for $n$;
$$n =\dfrac{\left(m+\frac{1}{2}\right)\lambda}{2t}$$
Plugging the given; and for minimum $n$, we use $m=0$
$$n=\dfrac{\left(0+\frac{1}{2}\right)675\times 10^{-9}}{2\times 125\times 10^{-9}}=\color{red}{\bf 1.35}$$
which is likely to be used.
