A soap film of refractive index 1.33 and thickness 380 nm stands in air…

Step 1. The path. $2nt = 2 \times 1.33 \times 380\ \text{nm} = 1010.8\ \text{nm}$.

Step 2. The bright wavelengths. $\lambda = \dfrac{2nt}{m + 1/2} = \dfrac{1010.8}{m + 1/2}$. For $m = 0$ it is $2022$ nm, outside the range. For $m = 1$ it is $673.9$ nm. For $m = 2$ it is $404.3$ nm. For $m = 3$ it is $289$ nm, below the range.

Step 3. The dark wavelength. $\lambda = \dfrac{2nt}{m} = \dfrac{1010.8}{m}$. For $m = 2$ it is $505.4$ nm. The others, $1011$ nm for $m = 1$ and $337$ nm for $m = 3$, are outside the visible range.