Answer
a) ${\bf 1.0}\;\rm mT$
b) ${\bf 0.16}\;\rm mT$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the magnetic field at a point at a distance $r$ from the center of a current-carrying wire is given by
$$B_{\rm wire}=\dfrac{\mu_0I_{\rm wire}}{2\pi r}$$
Plug the known;
$$B_{\rm wire}=\dfrac{(4\pi\times 10^{-7})(10)}{2\pi (2\times 10^{-3})}$$
$$B_{\rm wire}=\color{red}{\bf 1.0}\;\rm mT$$
$$\color{blue}{\bf [b]}$$
Recalling that the induced magnetic field inside a charging capacitor is given by
$$B=\dfrac{\mu_0 r}{2\pi R^2}\dfrac{dQ}{dt} $$
where $R$ is the radius of the capacitor's plate, and $dQ/dt=I_{\rm wire}$;
$$B=\dfrac{\mu_0 rI_{\rm wire}}{2\pi R^2} $$
Plug the known;
$$B=\dfrac{(4\pi\times 10^{-7})(2\times 10^{-3})(10)}{2\pi (0.5\times 10^{-2})^2} $$
$$B =\color{red}{\bf 0.16}\;\rm mT$$
