Answer
a) ${\bf 0.025}\;\rm A\cdot m^2$
b) ${\bf 1.48}\;\rm \mu T$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the on-axis field of a magnetic dipole is given by
$$\vec B_{\rm dipole}=\dfrac{\mu_0}{4\pi}\dfrac{2\vec\mu}{z^3}\tag 1$$
where $\vec\mu$ is the magnetic dipole, so the magnetic moment dipole magnitude is given by
$$ \mu=\dfrac{4\pi \vec B_{\rm dipole}z^3}{2\mu_0}$$
Plugging the known;
$$ \mu=\dfrac{4\pi (5\times 10^{-6})(0.1)^3}{2(4\pi \times 10^{-7})}$$
$$\mu=\color{red}{\bf 0.025}\;\rm A\cdot m^2$$
$$\color{blue}{\bf [b]}$$
Using (1),
$$ B_{\rm dipole}=\dfrac{(4\pi\times 10^{-7})}{4\pi}\dfrac{2(0.025)}{(0.15)^3} $$
$$ B_{\rm dipole}=\color{red}{\bf 1.48}\;\rm \mu T$$
