Answer
$43\;\rm cm$
Work Step by Step
We know that each light color will diffract by a different angle.
To find the width of the first order of the rainbow on the screen, we need to find $y_1$ for the 400 nm light and $y_1'$ for the 700 nm light.
Hence,
$$\Delta y=y_1'-y_1\tag 1$$
In a diffraction grating, we know that,
$$d\sin\theta_m=m\lambda$$
Hence, $\theta=\sin^{-1}\left[ \dfrac{m\lambda}{d} \right]$ where $d=1/N$ where $N$ is the number of slits in a unit of length.
So,
$$\theta=\sin^{-1}\left[ m\lambda N \right]\tag 2$$
Recalling that,
$$y_m=L\tan\theta_m$$
Plugging from (2);
$$y_m=L\tan\left( \sin^{-1}\left[ m\lambda N \right]\right)$$
Plugging into (1);
$$\Delta y=y_1'-y_1=L\tan\left( \sin^{-1}\left[ m\lambda' N \right]\right)-L\tan\left( \sin^{-1}\left[ m\lambda N \right]\right)$$
Plugging the known;
$$\Delta y =2\tan\left( \sin^{-1}\left[ (700\times 10^{-9}) (\frac{600}{1\times 10^{-3}}) \right]\right)-2\tan\left( \sin^{-1}\left[ (400\times 10^{-9}) (\frac{600}{1\times 10^{-3}}) \right]\right)$$
$$\Delta y=\color{red}{\bf0.43}\;\rm m$$
