Answer
$L = 52.5~nm$
Work Step by Step
Note that each ray reflects off an even number of mirrors, so the phase difference will be caused by the difference in path length.
We can measure each path length from the first red arrow, to the point that is a horizontal distance $L$ from the last mirror in each path.
The path length of ray 1 is $7L$
The path length of ray 1 is $3L$
The path length difference is $4L$
To be exactly out of phase, the path length difference must have the form $~~(m+\frac{1}{2})~\lambda$
We can let $m = 0$ to find the smallest value of $L$:
$4L = (m+\frac{1}{2})~\lambda$
$L = \frac{\lambda}{8}$
$L = \frac{420.0~nm}{8}$
$L = 52.5~nm$
