The wavelength increases since the path difference increases. The path length’s are

$$
l_1=sqrt{L^2+(d-y)^2} =14.0text{ km}; quad l_2=sqrt{L^2+(d+y)^2}=15.5text{ km}.
$$
This yields a path difference of
$$
Delta l= l_2-l_1 = 1.5text{ km}.
$$
We will determine $lambda$ from the condition of first destructive interference

$$
Delta L= frac{lambda}{2}Rightarrow lambda = 2Delta L = 3text{ km}.
$$

The path difference is
$$
Delta l=l_2-l_1=143text{ m} – 78text{ m} =65text{ m}.
$$
Note that

$$
Delta l=5frac{lambda}{2}=5times13
$$
which is odd integer times half of a wavelength so finally we conclude that destructive interference happens.

The path difference is
$$
Delta l=295text{ m}-161text{ m} = 134text{ m}.
$$
In order for constructive interference to happen $Delta l$ has to be a whole number times the wavelength. Obviously the largest wavelength that satisfies this is the one that is exactly equal to $Delta l$ i.e.

$$
lambda =Delta l= 134text{ m}.
$$