Answer
a) $2\;\rm \mu T$
b) $4\%$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the magnetic field of a long straight current wire is given by
$$B=\dfrac{\mu_0I}{2\pi d}$$
Plug the known;
$$B=\dfrac{(4\pi \times 10^{-7})(200)}{2\pi (20)}$$
$$B=\color{red}{\bf 2\times 10^{-6}}\;\rm T$$
$$\color{blue}{\bf [b]}$$
We can find the percentage of this wire's magnetic field relative to the earth’s magnetic field of 50 $\mu$T by dividing.
$$\dfrac{B_{\rm wire}}{B_{\rm Earth}}=\dfrac{2\mu}{50\mu}=0.04=\color{red}{\bf 4}\%$$
Thus the wire's magnetic field is just 4% of that of Earth.
