Answer
0.208 mm.
Work Step by Step
The value of $y_m$ is much smaller than R. Use the approximate expression $y_m=R\frac{m \lambda}{d}$.
$\Delta y =R\frac{\lambda}{d}$.
$$d=\frac{R\lambda}{\Delta y }=\frac{ (1.80m)(450\times10^{-9}m)}{3.90\times10^{-3}m}=2.08\times10^{-4}m$$
