The phase of a reflected wave changes for $180^circ$ in respect to the one that is incident only when the reflection occurs on a medium (medium 2 on the figure) that has higher index of refraction than the medium in which both the incident and the reflected wave travel (medium 1 on the figure) so here we conclude that the index of medium 2 has to be higher than $1.35$.

Ray 1 reflects on the material with lower index of refraction than that of the material it travels in so there is no phase change. Ray 2 reflects on the material that has higher index of refraction than that of the medium it travels in so it changes its’ phase for $180^circ$ meaning that the phase shift of ray 1 due to reflection is less than that of the ray 2.

Here the light reflecting on the upper surface (glass-air boundary) experiences no phase shift while the light reflecting on the lower surface (air-glass boundary) experiences a shift of $180^circ$ so here the condition for constructive interference reads

$$
2n_{air}t = left(m+frac{1}{2}right)lambda.
$$
Putting $n_{air} = 1$ and $m=250$ we get

$$
2t=250.5lambdaRightarrow lambda = frac{2t}{250.5} = frac{2times5.1times10^{-5}text{ m}}{250.5} = 407text{ nm}.
$$

Here we have to apply the condition for destructive interference which reads

$$
2nt = mlambda.
$$

Putting $m=1$ we have

$$
2nt=lambdaRightarrow n=frac{lambda}{2t} = frac{523text{ nm}}{2times 195text{ nm}} =1.34.
$$

The green light is absent from the reflection because it interferes destructively. We can write the condition for destructive interference

$$
2n_{oil}t = mlambda
$$
yielding for the thickness

$$
t=frac{mlambda}{2n_{oil}}.
$$
We see that the thickness is minimal when we put $m=1$ so finally we obtain
$$
t=frac{lambda}{2n_{oil}} = frac{521text{ nm}}{2times1.38} = 189text{ nm}.
$$