Answer
2.53 mm.
Work Step by Step
The values of y are much smaller than R. Use the approximate expression $y_m=R\frac{m \lambda}{d}$.
$\Delta y_1=R\frac{\Delta \lambda}{d}$
$$ =(4.00m)\frac{ (190\times10^{-9}m)}{0.300\times10^{-3}m}=2.53\times10^{-3}m$$
