Answer
a) ${\bf 20}\;\rm A$
b) ${\bf 1.6}\;\rm mm$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the magnetic field of a long straight current wire is given by
$$B=\dfrac{\mu_0I}{2R}$$
Hence, the current needed is given by
$$I =\dfrac{2R B }{\mu_0}$$
Plug the known;
$$I =\dfrac{2(0.5\times 10^{-2})(2.5\times 10^{-3}) }{4\pi\times 10^{-7}}$$
$$I_{\rm loop}=\color{red}{\bf 20}\;\rm A$$
$$\color{blue}{\bf [b]}$$
We know that the magnetic field of a long straight current wire is given by
$$B=\dfrac{\mu_0I}{2\pi d}$$
Hence, the distance needed from the long thin wire is given by
$$d=\dfrac{\mu_0I}{2\pi B}$$
For current of $I=20\;\rm A$;
$$d=\dfrac{(4\pi \times 10^{-7})(20)}{2\pi B}$$
$$d=\dfrac{(40 \times 10^{-7}) }{ B}=\dfrac{(40 \times 10^{-7}) }{ (2.5\times 10^{-3})}$$
$$d =\color{red}{\bf 1.6}\;\rm mm$$
