Answer
$h=1.66 \times 10^3 \mathrm{~km} .$
Work Step by Step
At a distance $r$ from the center of the Earth, the magnitude of the magnetic field is given by
$$
B=\frac{\mu_0 \mu}{4 \pi r^3} \sqrt{1+3 \sin ^2 \lambda_w},
$$
where $\mu$ is the Earth's dipole moment and $\lambda_m$ is the magnetic latitude. The ratio of the field magnitudes for two different distances at the same latitude is
$$
\frac{B_2}{B_1}=\frac{r_1^3}{r_2^3} .
$$
With $B_1$ being the value at the surface and $B_2$ being half of $B_1$, we set $r_1$ equal to the radius $R_e$ of the Earth and $r_2$ equal to $R_e+h$, where $h$ is altitude at which $B$ is half its value at the surface. Thus,
$$
\frac{1}{2}=\frac{R_c^3}{\left.\mid R_e+h\right]^3} .
$$
Taking the cube root of both sides and solving for $h$, we get
$
h=\left(2^{1 / 3}-1\right) R_e=\left(2^{1 / 3}-1\right)(6370 \mathrm{~km})=1.66 \times 10^3 \mathrm{~km} .
$
