Answer
$500\;\rm nm$
Work Step by Step
In the given figure, we are given the distance of the first-order bright fringe at $m=1$ which is $y_1=43.6$ cm.
So, we can find the angle of diffraction $\theta$ at $m=1$ by using $d\sin\theta_m=m\lambda$. Hence,
$$\lambda=d\sin\theta_1$$
where $\theta_1=\tan^{-1}\left[ \dfrac{y_1}{L}\right]$, and $d=\dfrac{1}{N}$.
$$\lambda=\dfrac{1}{N}\;\sin\left(\tan^{-1}\left[ \dfrac{y_1}{L}\right]\right)$$
Plugging the known;
$$\lambda=\dfrac{10^{-3}}{800}\;\sin\left(\tan^{-1}\left[ \dfrac{43.6}{100}\right]\right)=\bf 4.995\times 10^{-7}\;\rm m$$
$$\lambda=\color{red}{\bf 500}\;\rm nm$$
