#### Answer

(a) $\mu = 0.025~A~m^2$
(b) $B = 1.5~\mu T$

#### Work Step by Step

(a) We can find the magnetic moment:
$B = \frac{\mu_0~\mu}{2\pi~r^3}$
$\mu = \frac{2\pi~r^3~B}{\mu_0}$
$\mu = \frac{(2\pi)~(0.10~m)^3~(5.0\times 10^{-6}~T)}{4\pi\times 10^{-7}~T~m/A}$
$\mu = 0.025~A~m^2$
(b) We can find the on-axis field strength:
$B = \frac{\mu_0~\mu}{2\pi~r^3}$
$B = \frac{(4\pi\times 10^{-7}~T~m/A)(0.025~A~m^2)}{(2\pi)~(0.15~m)^3}$
$B = 1.5\times 10^{-6}~T$
$B = 1.5~\mu T$