Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Recalling that the magnetic dipole moment of a loop is given by
$$\mu =IA_{\rm loop}=\pi R_{\rm loop}^2 I$$
Hence,
$$I=\dfrac{\mu}{\pi R_{\rm loop}^2}$$
But what diameter should we use? The total diameter of 4000-km or the 3000-km diameter as the author told us? Since we are dealing with an imaginary model, we can use the 3000-km diameter.
Plug the given;
$$I=\dfrac{(8\times 10^{22})}{\pi (1500\times 10^3)^2}$$
$$I=\color{red}{\bf 1.13\times 10^{10}}\;\rm A$$
$$\color{blue}{\bf [b]}$$
The current density is given by
$$J=\dfrac{I}{A_{\rm wire}}=\dfrac{I}{\pi R_{\rm wire}^2}$$
Plug the known;
$$J= \dfrac{1.13\times 10^{10}}{\pi (500\times 10^3)^2}$$
$$J=\color{red}{\bf 0.0144}\;\rm A/m^2$$
$$\color{blue}{\bf [c]}$$
Let's assume that the 1 mm-diameter wire has a current density of $J'$, so
$$\dfrac{J}{J'}=\dfrac{0.0144}{\dfrac{1}{\pi (0.5\times 10^{-3})^2}}=\bf 1.13\times 10^{-8}$$
Thus,
$$J=1.13\times 10^{-8} J'$$
We can see that the wire inside the Earth's core is too large than we can handle but the current density is too small.
This means that the current density of this giant loop wire is negligible relative to this tinny wire.
