Answer
2.52 m.
Work Step by Step
We see from Figure 24-21 and equation 24-3b that the center of the first bright fringe is characterized by $sin \theta\approx \frac{3}{2}\frac{\lambda}{D}$.
Find the angle.
$$ \theta=sin^{-1} \frac{3}{2}\frac{\lambda}{D}=sin^{-1} \frac{3}{2}\frac{620\times10^{-9}m}{3.80\times10^{-6}m}=14.17^{\circ}$$
Now find the distance of the fringe from the central maximum, on the screen. Multiply the distance to the screen by the tangent of the above angle.
$$y=L\;tan\theta=(10.0m)tan(14.17^{\circ})=2.52m$$
