Answer
a) ${\bf 5.73}\;\rm \mu A$
b) ${\bf2.88\times 10^{-8}}\;\rm A\cdot m^2$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We can model this problem as a current loop which we need to find its magnetic field strength at its center.
We know that the magnitude of the magnetic field at some point on the axis of a current loop is given by
$$B=\dfrac{\mu_0 I R^2}{2(z^2+R^2)^{3/2}} $$
At the center of the loop where $z=0$, so
$$B=\dfrac{\mu_0 I R^2}{2R^3} $$
$$B=\dfrac{\mu_0 I }{2R } $$
We need to find the current, so
$$I=\dfrac{ 2R B}{\mu_0}$$
Plug the known;
$$I=\dfrac{ 2(0.04)(90\times 10^{-12})}{(4\pi\times 10^{-7})}$$
$$I=\color{red}{\bf 5.73}\;\rm \mu A$$
$$\color{blue}{\bf [b]}$$
Thus the heart’s magnetic dipole moment is the same as the loop,
$$\mu =AI=\pi R^2 I$$
Plug the known;
$$\mu =\pi (0.04)^2 (5.73\times 10^{-6})$$
$$\mu=\color{red}{\bf2.88\times 10^{-8}}\;\rm A\cdot m^2$$
