Answer
$ 666\;\rm line/mm$
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 $y_m=L\tan\theta $.
Hence,
$$\theta=\tan^{-1}\left[ \dfrac{y_1}{L} \right]\tag 1$$
And we also know that
$$d\sin\theta_m=m\lambda$$
So, at $m=1$,
$$d=\dfrac{\lambda}{\sin\theta_1}$$
Recall that we need to find the number of lines per millimeter, not $d$, where $N=1/d$
Hence,
$$N=\dfrac{{\sin\theta_1}}{\lambda}$$
Plugging from (1);
$$N=\dfrac{{\sin\left(\tan^{-1}\left[ \dfrac{y_1}{L} \right]\right)}}{\lambda}$$
Plugging the known;
$$N=\dfrac{{\sin\left(\tan^{-1}\left[ \dfrac{43.6}{100} \right]\right)}}{600\times 10^{-9}}=\bf 6.66\times 10^5\;\rm line/m$$
$$N=\color{red}{\bf 666}\;\rm line/mm$$
