What is the approximate earth's dipole moment, if the earth's magnetic field at the Equator is approximately 0.5 G and diameter is $15.0 × 10^6m$?

$2.1 × 10^{23} Am^2$

The correct answer is Option (3) → $2.1 × 10^{23} Am^2$

Earth's magnetic dipole moment is,

$B=\frac{μ_0M}{4πr^3}$

$M=\frac{(0.5×10^{-4})×4π(1.5×10^6)^3}{4π×10^{-7}}$

$=2.1×10^{23} Am^2$