Answer
$799.85\;\rm line/mm$
Work Step by Step
To find how many lines per mm, we need to find $d$.
We know, in a diffraction grating, that the angle of the bright fringes is given by
$$d\sin\theta_m=m\lambda$$
And hence, for the first order bright fringe where $m=1$
$$d\sin\theta_1= \lambda\tag 1$$
where the angle is given by $y_1=L\tan\theta_1$.
So
$$\theta_1=\tan^{-1}\left[ \dfrac{y_1}{L}\right]$$
where $y_1$ is half the distance between the two first-order bright fringes; $y_1=\dfrac{\Delta y}{2}$
Thus,
$$\theta_1=\tan^{-1}\left[ \dfrac{\Delta y}{2L}\right]$$
Plugging into (1);
$$d\sin\left(\tan^{-1}\left[ \dfrac{\Delta y}{2L}\right]\right)= \lambda $$
we are given $\lambda$ and $\Delta y$ in the given table. And $L$ is constant and it is 150 cm.
$$d\sin\left(\tan^{-1}\left[ \dfrac{\Delta y}{2(1.50)}\right]\right)= \lambda $$
$$d\sin\left(\tan^{-1}\left[ \dfrac{\Delta y}{3}\right]\right)= \lambda $$
$$ \sin\left(\tan^{-1}\left[ \dfrac{\Delta y}{3}\right]\right)=\dfrac{1}{d} \lambda $$
Recalling that $d=1/N$, so
$$ \boxed{\sin\left(\tan^{-1}\left[ \dfrac{\Delta y}{3}\right]\right)=N \lambda }$$
Now we have a straight-line formula ($y=mx+b$) where $y=\sin\left(\tan^{-1}\left[ \dfrac{\Delta y}{3}\right]\right)$, $m={\rm slope}=N$, and $x=\lambda$.
We need to use the given table to find the following dots in the table below.
\begin{array}{|c|c|c|c|}
\hline
x=\lambda\;\rm (nm)& y=\sin\left(\tan^{-1}\left[ \dfrac{\Delta y}{3}\right]\right) \\
\hline
430&0.34315 \\
\hline
480 & 0.38566 \\
\hline
530&0.422389 \\
\hline
580 & 0.464139 \\
\hline
630 & 0.50258 \\
\hline
680 & 0.544595 \\
\hline
\end{array}
From the graph below, it is obvious that the best-fit line slope is
$$N=\color{red}{\bf 799.85}\;\rm line/mm$$
