Answer
a) ${\bf 3.14\times 10^{-4}}\;\rm A\cdot m^2$
b) ${\bf 5.02\times 10^{-7}}\;\rm T$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the magnetic dipole moment of the superconducting ring is given by
$$\mu_{\rm loop}= \pi R^2 I$$
Plugging the known;
$$\mu_{\rm loop}= \pi (1\times 10^{-3})^2(100)$$
$$\mu_{\rm loop}=\color{red}{\bf 3.14\times 10^{-4}}\;\rm A\cdot m^2$$
$$\color{blue}{\bf [b]}$$
We know that the on-axis magnetic field of the superconducting ring is given by
$$ B_{\rm loop}=\dfrac{\mu_0 IR^2}{2(R^2+z^2)^{3/2}}$$
Plug the known;
$$ B_{\rm loop}=\dfrac{(4\pi\times 10^{-7}) (100)(0.001)^2 }{2(0.001^2+0.05^2)^{3/2}} $$
$$ B_{\rm loop}=\color{red}{\bf 5.02\times 10^{-7}}\;\rm T$$
