Answer
$L = 478~nm$
Work Step by Step
Since $n_2 \gt n_1$ and $n_2 \gt n_3$, there is no phase shift due to reflection.
Since we are looking for a minimum, the path length difference must be $~~(m+\frac{1}{2})~\frac{\lambda}{n_2}~~$, where $m$ is an integer.
We can find the required thickness $L$:
$2L = \frac{(m+\frac{1}{2})~\lambda}{n_2},$ where $m$ is an integer
$L = \frac{(m+\frac{1}{2})~\lambda}{2~n_2}$
$L = \frac{(m+\frac{1}{2})~(612~nm)}{(2)~(1.60)}$
$L = (m+\frac{1}{2})~(191.25~nm)$
$L= 95.6~nm, 287~nm, 478~nm,...$
Since we are looking for the third least thickness, $~~L = 478~nm$
