Answer
a) $0.52\;\rm mm$
b) $0.074^\circ$
c) $1.3\;\rm m$
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since it is similar to the circular aperture,
$$w=\dfrac{2.44\lambda L}{D}$$
And since we need the central maximum equals the diameter of the pinhole,
Thus,
$$D^2=2.44\lambda L$$
$$D=\sqrt{2.44\lambda L}$$
Plugging the known;
$$D=\sqrt{2.44(550\times 10^{-9})(20\times 10^{-2})}=\bf 5.2\times 10^{-4}\;\rm m$$
$$D=\color{red}{\bf 0.52}\;\rm mm$$
$$\color{blue}{\bf [b]}$$
We can use the formula of,
$$\alpha=\theta_1=\dfrac{1.22\lambda}{D}$$
to find the angle between the two distant sources that can be resolved.
Plugging the known;
$$\alpha =\dfrac{1.22(550\times 10^{-9}) }{ 5.2\times 10^{-4}}=\bf 1.29\times 10^{-3} \;\rm rad=\color{red}{\bf0.074^\circ}$$
$$\color{blue}{\bf [c]}$$
The distance between two street lights 1 km away that can barely be resolved is given by
$$x=L\alpha=(1000)(1.29\times 10^{-3})=\color{red}{\bf 1.3}\;\rm m$$
