The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of $$20$$ cm from its center is $$1.5 \times 10^{-5}$$ T m. The magnetic moment of the dipole is _______ A m$$^2$$. (Given: $$\frac{\mu_0}{4\pi} = 10^{-7}$$ T m A$$^{-1}$$)

Magnetic potential on axis: $$V = \frac{\mu_0}{4\pi}\frac{M\cos\theta}{r^2}$$. On axis, $$\theta = 0$$:

$$V = \frac{\mu_0 M}{4\pi r^2} = 10^{-7} \times \frac{M}{(0.2)^2} = 10^{-7} \times 25M$$

$$1.5 \times 10^{-5} = 25 \times 10^{-7} M$$. $$M = \frac{1.5 \times 10^{-5}}{25 \times 10^{-7}} = 6$$ Am².

The answer is $$\boxed{6}$$.