Step 1: Build the equatorial field from two pole fields, instead of quoting the bar-magnet formula directly.
Picture the bar magnet as two poles, $+m_p$ and $-m_p$, separated by a small distance $2l$. At a point on the perpendicular bisector, a distance $r$ from the centre, each pole is actually $\sqrt{r^2+l^2}$ away and contributes a field of size $\frac{\mu_0}{4\pi}\frac{m_p}{r^2+l^2}$ along the line joining pole to point.
Step 2: Add the axial components of the two pole fields.
The perpendicular components cancel by symmetry, and the axial components add, each picking up a factor $\frac{l}{\sqrt{r^2+l^2}}$ from the triangle geometry: \[ B = 2 \times \frac{\mu_0}{4\pi}\frac{m_p}{r^2+l^2} \times \frac{l}{\sqrt{r^2+l^2}} = \frac{\mu_0}{4\pi}\frac{2m_p l}{(r^2+l^2)^{3/2}} \]
Step 3: Use the short-magnet limit ($l \ll r$) and $M = 2m_p l$.
\[ B \approx \frac{\mu_0}{4\pi}\frac{M}{r^3} \]
Step 4: Substitute the given numbers.
\[ M = \frac{Br^3}{\mu_0/4\pi} = \frac{5 \times 10^{-6} \times (0.2)^3}{10^{-7}} \]
Step 5: Work through the arithmetic.
\[ (0.2)^3 = 8 \times 10^{-3},\qquad 5 \times 10^{-6} \times 8 \times 10^{-3} = 4 \times 10^{-8} \] \[ M = \frac{4 \times 10^{-8}}{10^{-7}} = 0.4\ \text{J T}^{-1} \]
Final Answer:
\[ \boxed{0.4\ \text{J T}^{-1}} \]