Answer
Choice A.
Work Step by Step
For an interference maximum, $2d sin \theta=m \lambda/n$.
Solve for d using the given numbers, using n $\approx$ 1.3 for water.
$d =\frac{m \lambda}{2n sin \theta}=\frac{1 (650 nm)}{2(1.3) sin 39^{\circ}}\approx 390 nm$
