Answer
$\dfrac{\sqrt{2}\;d}{2}$
Work Step by Step
We are given that the distance to the screen is $L$, and the position of the first-order bright fringe is $y_1=L$.
We know, in a diffraction grating, that
$$d\sin\theta_m=m\lambda$$
So at $m=1$,
$$\lambda=d\sin\theta_1$$
where $\theta_1=\tan^{-1}\left[\dfrac{y_1}{L}\right]=\tan^{-1}\left[\dfrac{L}{L}\right]=45^\circ$
Hence,
$$\lambda=d\sin45^\circ$$
$$\boxed{\lambda=\dfrac{\sqrt{2}\;d}{2} }$$
