Answer
The height above Earth's surface is about $6300~m$
Work Step by Step
Since three-quarters of the muons decay, the time period is two half-lives. In the muon's frame, $t = (2)(1.5~\mu s) = 3.0~\mu s$
We can find the height $h_m$ in the muon's frame:
$h_m = v~t$
$h_m = (0.990)(3.0\times 10^8~m/s)(3.0\times 10^{-6}~s)$
$h_m = 891~m$
We can find the height $h_0$ in the Earth's reference:
$h_m = \frac{h_0}{\gamma}$
$h_0 = h_m~\gamma$
$h_0 = \frac{h_m}{\sqrt{1-\frac{v^2}{c^2}}}$
$h_0 = \frac{891~m}{\sqrt{1-\frac{(0.990)^2}{c^2}}}$
$h_0 = \frac{891~m}{\sqrt{1-0.9801}}$
$h_0 = 6300~m$
The height above Earth's surface is about $6300~m$.
