Answer
$L=161nm$
Work Step by Step
We know that
$2L=(m+\frac{1}{2})(\frac{\lambda}{n_2})$
This can be rearranged as:
$L=(m+\frac{1}{2})(\frac{\lambda}{2n_2})$
For $m=1$ and plugging in the known values to obtain:
$L=(1+\frac{1}{2})(\frac{342}{2\times 1.59})=161nm$
