Answer
$914.4 *10^{6}$ gratings per milimetre
Work Step by Step
The question establishes that the light has a wavelength spectrum from $\lambda1=430nm$ to $\lambda2=680nm$. The constraint set is that the first order diffraction spectrum must range over an angular width of $20.0$ degrees. The angle at which the maxima of a particular wavelength is formed is given by $$dsin\theta=m\lambda$$
where $d$ is the grating spacing, $\theta$ is the angular position at which the maxima corresponding to the wavelength $\lambda$ forms. $m$ is the order of the diffraction maxima formed.
As we can see that for a fixed order and diffraction spacing the larger the wavelength, the larger the angle at which it is formed. In the question we deal with $m=1$ the first order of diffraction and the angles corresponding to $\lambda1$ ,$\theta1$ and $\lambda2$ ,$\theta2$ must have an angular difference of $20$ degrees, so as to span the entire spectrum in this width.
$$\theta2-\theta1=20^{\circ}$$
This the equations we develop based on this information are
$$d*sin\theta1=430nm$$
$$d*sin\theta2=680nm$$
We know that $d$ can only have one value. Hence we can equate the two equations and get
$$\frac{430}{sin\theta1}=\frac{680}{sin\theta2}$$
We have two unknowns in this equation, which makes it tough to solve. However the key here is to realize that we have been provided with a constraint on the angles in the question, $\theta2-\theta1=20^{\circ}$. Hence, we can write $\theta2=\theta1+20^{\circ}$. Thus the equality can now be written as
$$\frac{430}{sin\theta1}=\frac{680}{sin(\theta1+20^{\circ})}$$
This equation is now solvable, using the identity $sin(A+B)=sinAcosB+cosAsinB$
Thus,
$$\frac{430}{sin\theta1}=\frac{680}{sin\theta1cos20+cos\theta1sin20}$$
$$=>(sin\theta1cos20+cos\theta1sin20)(430)=(sin\theta1)(680)$$
$$=>sin\theta1cos20(430)+cos\theta1sin20(430)=sin\theta1(680)$$
Now we divide both sides of the equation by $sin\theta1$ and obtain
$$cos20(430)+cot\theta1sin20(430)=(680)$$
Simplifying the trigonometric terms and the equation, we obtain
$$404.06+147.06cot\theta1=680$$
Thus,
$$cot\theta1=1.876$$
$$\theta1=28.05^{\circ}$$
We have this now found the angle to which the first order maxima formed by $\lambda1$ corresponds to. We can plug that into $d*sin\theta1=430nm$ to find the value of $d$.
Hence we get $$d=\frac{430}{sin(28.05)}=914.4nm$$
Therefore we have found that the spacing between the gratings is $914.4 *10^{-9} m$. This allows us to conclude that in one $1m$ there will be $\frac{1}{914.4 *10^{-9}}=914.4 *10^{9}$ gratings. Thus, in one $1mm$, we have $\frac{914.4 *10^{9}}{10^{3}}=914.4 *10^{6}$ gratings.
