Answer
a) ${\bf 3.93\times 10^{-4}}\;\rm J/m^3 $
b) ${\bf 3.14}\;\rm A $
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the magnetic energy density is given by
$$u_B=\dfrac{B^2}{2\mu_0}\tag 1$$
where the magnetic field strength at the center of a current loop is given by
$$B=\dfrac{\mu_0 I}{2r}$$
Plug into (1),
$$u_B=\dfrac{1}{2\mu_0}\left[\dfrac{\mu_0 I}{2r}\right]^2 $$
$$u_B=\dfrac{1}{2\mu_0}\left[\dfrac{\mu_0^2 I^2}{4r^2}\right] $$
$$u_B= \dfrac{\mu_0 I^2}{8r^2} $$
Plug the known;
$$u_B= \dfrac{(4\pi \times 10^{-7})(1)^2}{8(0.02)^2} $$
$$u_B=\color{red}{\bf 3.93\times 10^{-4}}\;\rm J/m^3 $$
$$\color{blue}{\bf [b]}$$
We know that the magnetic field strength at a distance $r$ from a straight current wire is given by
$$B=\dfrac{\mu_0 I}{2\pi r}$$
Plug into (1),
$$u_B=\dfrac{1}{2\mu_0}\left[\dfrac{\mu_0 I}{2\pi r}\right]^2 $$
$$u_B=\dfrac{1}{2\mu_0}\left[\dfrac{\mu_0^2 I^2}{4\pi ^2r^2}\right] $$
$$u_B= \dfrac{\mu_0 I^2}{8\pi ^2r^2} $$
Solving for $I$;
$$I=\sqrt{\dfrac{8\pi ^2r^2u_B}{\mu_0}}$$
Plug the known;
$$I=\sqrt{\dfrac{8\pi ^2(0.02)^2 (3.93\times 10^{-4})}{(4\pi \times 10^{-7})}}$$
$$I=\color{red}{\bf 3.14}\;\rm A $$