(a) On axial position
$B_{a} = \frac{\mu_{0}}{4\pi}\frac{2Mr}{\left( r^{2} - l^{2} \right)^{2}};if\ l < \ < r,\ then\ B_{a} = \frac{\mu_{0}}{4\pi}\frac{2M}{r^{3}}$
$B_{a} = \frac{10^{- 7}\ \times \ 2\ \times \ 10}{(0.1)^{3}} = 2 \times 10^{- 3}\ T\ \ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because z = r\rbrack$