Answer
$\lambda = 442~nm$
Work Step by Step
Note that light in the visible range has a wavelength of approximately $400~nm$ to $700~nm$
Let $n_1 = 1.00$ be the index of refraction of air
Let $n_2 = 1.20$ be the index of refraction of kerosene
Let $n_3 = 1.30$ be the index of refraction of water
Since $n_3 \gt n_2$, there is a phase shift of $0.5~\lambda$ due to reflection from the bottom surface of the kerosene layer.
Since $n_2 \gt n_1$, there is no phase shift due to reflection from the top surface of the kerosene layer.
Since we are looking for a maximum (constructive interference), the path length difference must be $~~(m+0.5)~\frac{\lambda}{n_2}~~$, where $m$ is an integer.
We can find the required wavelength $\lambda$:
$2L = \frac{(m+0.5)~\lambda}{n_2},$ where $m$ is an integer
$\lambda = \frac{2~L~n_2}{(m+0.5)}$
$\lambda = \frac{(2)~(460~nm)~(1.20)}{(m+0.5)}$
$\lambda = \frac{1104~nm}{(m+0.5)}$
$\lambda = 2208~nm, 736~nm, 442~nm, 315~nm,...$
Since $442~nm$ is in the visible range, $~~\lambda = 442~nm$
