Calculating Dipole Magnetic Moment Given Magnetic Field Strength I am trying to figure out how to solve the following:

The Earth's magnetic field can be represented, in a first approximation, by a magnetic dipole placed in the Earth's center, at least up to
  distances of a few Earth radii ($R_E$)·
Using the fact that, at one of the magnetic poles, the field has a magnitude of approximately 0.5 gauss near the surface, calculate the dipole magnetic moment, $\mu$.

How does one go about calculating the dipole magnetic moment given the magnetic field strength at one the Earth's poles?
It appears as though applying the equation
$
\vec{B} = \frac{\mu_0}{4\pi r^3}(3(\vec{\mu}\cdot \hat{r})\hat{r}-\vec{\mu}) 
$
will be useful but I am not sure if that is the correct approximation to make here.
 A: Indeed, we can approximate the Earth's magnetic field as a dipole and apply:
$
\vec{B} = \frac{\mu_0}{4\pi r^3}(3(\vec{\mu}\cdot \hat{r})\hat{r}-\vec{\mu}) 
$
Where we know via the problem statement what the magnetic field strength at the North pole is, $|B_{N_{pole}}|$, and that the radius at the pole will be one earth radii, $R_{Earth}$, since it is assumed in the problem that the dipole approximation should be made from the Earth's center. Further, $\vec{\mu}$ points from South to North, and thus at the North pole we have
$\vec{\mu}\cdot \hat{r} = |\vec{\mu}||\hat{r}|\cos(0^\circ)= |\vec{\mu}|$
, so applying the magnetic field of a dipole approximation at the North pole with $|\vec{B}_{N_{pole}}|$ we have:
$\begin{align}
|\vec{B}_{N_{pole}}| &= \Big|\frac{\mu_0}{4\pi R_{Earth}^3}(3(\vec{\mu}\cdot \hat{r})\hat{r}-\vec{\mu})\Big|\\
\therefore \frac{4\pi R_{Earth}^3|\vec{B}_{N_{pole}}|}{\mu_0} &= \Big|3|\vec{\mu}| \hat{r}-\vec{\mu}\Big| = |\vec{\mu}|\Big|3 \hat{r}-\hat{r}\Big| = 2|\vec{\mu}|\\
|\vec{\mu}| &= \frac{4\pi R_{Earth}^3|\vec{B}_{N_{pole}}|}{2\mu_0}\\
|\vec{\mu}| &= \frac{2\pi R_{Earth}^3|\vec{B}_{N_{pole}}|}{\mu_0}
= \frac{2\cdot \pi \cdot (6.38 \times 10^6 m)^3\cdot (0.5 \times 10^{-4} T)}{(1.256 \times 10^{-6} NA^{-2})}\\
\therefore |\vec{\mu}| &\approx 6.48 \times 10^{22} Am^2
\end{align}$
Some insights were gleaned from this source: http://geophysics.ou.edu/solid_earth/notes/mag_earth/earth.htm
