Answer
a) ${\bf 6.16\times 10^{-15}}\;\rm T$
b) ${\bf 6.3\times 10^8}\;\rm A$
Work Step by Step
$$\color{blue}{\bf [a]}$$
Assuming that the dimensions of the loop are much smaller than the radius of the Earth since the author told us to assume that it is deep inside the Earth, so the on-axis magnetic field strength at 50 cm is given by
$$ B =\dfrac{2\mu_0 \mu }{4\pi z^3}$$
where $z=R_E$ where $R_E$ is the radius of the Earth.
Plug the known;
$$ B =\dfrac{2(4\pi\times 10^{-7}) (8\times 10^{22})}{4\pi (6.38\times 10^6)^3}$$
$$B=\color{red}{\bf 6.16\times 10^{-15}}\;\rm T$$
$$\color{blue}{\bf [b]}$$
We know that the magnetic dipole moment of the loop is given by
$$\mu =AI$$
where $A$ is the cross-sectional area of a loop and our loop here has a radius equal to the Earth's radius. so $A=\pi R_E^2$
Hence,
$$I=\dfrac{\mu }{A}=\dfrac{\mu}{\pi R_E^2}$$
Plug the known;
$$I= \dfrac{(8\times 10^{22})}{\pi (6.38\times 10^6)^2}$$
$$I=\color{red}{\bf 6.3\times 10^8}\;\rm A$$
