A short bar magnet has a magnetic moment of 4.8 A m²…

Step 1. On the axis. $B = \dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3} = 10^{-7}\ \text{T m/A} \times \dfrac{2 \times 4.8\ \text{A m}^2}{(0.52\ \text{m})^3} = \dfrac{9.6 \times 10^{-7}}{0.1406}\ \text{T} = 6.83 \times 10^{-6}\ \text{T} = 6.83\ \mu\text{T}$.

Step 2. On the equatorial line. $B = \dfrac{\mu_0}{4\pi}\dfrac{m}{r^3} = 3.41\ \mu\text{T}$, half of the axial value.

Step 3. The distance. We want $r$, so $r^3 = \dfrac{\mu_0}{4\pi}\dfrac{2m}{B} = \dfrac{10^{-7}\ \text{T m/A} \times 9.6\ \text{A m}^2}{4.0 \times 10^{-5}\ \text{T}} = 0.0240\ \text{m}^3$. Then $r = 0.288\ \text{m} = 28.8$ cm.