Here we have green light of wavelength $lambda=5.46times10^{-7}text{ m}$ passing through the slit that is $w=8times 10^{-5}text{ m}$ and appears on the screen at the distance $L=6.8times 10^{-1}text{ m}$. We need to find width of a band.

$$
begin{align}
2cdotlambda&=frac{2cdot xcdot w}{L} \
2cdot x&=frac{2cdotlambdacdot L}{w} \
2cdot x&=frac{2cdot5.46times10^{-7}text{ m}cdot 6.8times 10^{-1}text{ m}}{8times 10^{-5}m} \
&boxed{2cdot x=9.28times 10^{-3}m}
end{align}
$$

$$
2cdot x=9.28times 10^{-3}m
$$

We can use approximation

$$
begin{align}
sin{theta} approx tan{theta}
end{align}
$$

From the definition of the wavelength from a diffraction grating and basic geometry and with using approximation (1) we can consider

$$
begin{align}
lambda approx frac{xcdot}{L}
end{align}
$$

We can see that dot spacing $x$ and slit spacing $d$ are inversely proportional which means if the dots from the first pattern are greater dot spacing, more lines per millimeter will be on that grating.

In this problem we have to find the minimum separation between stars $x_{obj}$ if we know the diameter of the $Hubble$ $Space$ $Telescope$ $D=2.4text{m}$, the position of the Sirius system $L_{obj}=8.44text{ly}$ and the wavelength $lambda=5.5times10^{-7}text{ m}$. We will use The Rayleigh criterion to solve this problem.

$$
begin{align}
{x_{obj}}=frac{1.22cdotlambdacdot L_{obj}}{D}
end{align}
$$

We need to covert light years into meters to solve the equation.

It is know that one light year is equal to $9.4605284times10^{15}text{m}$. So for $L$ we have $L_{obj}=7.98times10^{16}text{m}$. Then we have

$$
begin{align}
{x_{obj}}&=frac{1.22cdot5.5times10^{-7}text{ m}cdot 7.98times10^{16}text{m}}{2.4text{m}} \
&boxed{x_{obj}=2.23times10^{10}text{m}}
end{align}
$$

$$
x_{obj}=2.23times10^{10}text{m}
$$