Answer
See the detailed answer below.
Work Step by Step
$$\bf a)$$ As we see in the figure below, we have two reflected rays one of them experiencing a 180$^\circ$ phase change since the water has an index of refraction than that of the air.
So for constructive interference for one phase change and two reflected rays,
$$2d=\left(m+\frac{1}{2}\right)\dfrac{\lambda}{n_{\rm film}}$$
where $d$ is the thickness of the film.
So,
$$\lambda_{\rm C}=\dfrac{2dn_{\rm film}}{m+\frac{1}{2}}$$
$$\boxed{\lambda_{\rm C}=\dfrac{2.66d}{m+\frac{1}{2}}}\tag {$m=0,1,2,3,...$}$$
$$\bf b)$$ For $d=390$ nm,
At $m=0$,
$$\lambda=\dfrac{2.66(390)}{0+\frac{1}{2}}=2075\;\rm nm$$
which is not visible light.
At $m=1$,
$$\lambda=\dfrac{2.66(390)}{1+\frac{1}{2}}=\color{red}{\bf 692}\;\rm nm$$
which is $\bf RED-color$ light.
At $m=2$,
$$\lambda=\dfrac{2.66(390)}{2+\frac{1}{2}}=\color{red}{\bf 415}\;\rm nm$$
which is $\bf VIOLET-color$ light.
At $m=3$,
$$\lambda=\dfrac{2.66(390)}{3+\frac{1}{2}}=296\;\rm nm$$
which is not visible light.
