Answer
See the detailed answer below.
Work Step by Step
First, let's assume that the Earth's frame is $\rm S$ and the muon’s frame is $\rm S'$ where the speed of $\rm S'$ relative to $\rm S$ is $v\approx c$ (We know that the speed of a muon is $0.9997c $).
The time it takes the muon to hit the ground in $\rm S$ frame is then given by
$$t=\dfrac{L}{c}=\dfrac{60\times 10^3}{3\times 10^8}=\bf 200\;\rm \mu s$$
where we know that the half-lifetime of a muon is only 1.5 $\mu$s.
Recalling that the muon speed is close to the speed of light which means that the distance traveled by the muon in $\rm S'$ frame is not 60,000 m, it is given by
$$L'=L\sqrt{1-\frac{v^2}{c^2}}$$
Plug the known;
$$L'=60,000\sqrt{1-\frac{0.9997^2c ^2}{c^2}}=\bf 1.47\;\rm km$$
Now we can find the time $t'$ of $\rm S'$ that makes the muon able to reach the ground.
$$t'=\dfrac{L'}{c}=\dfrac{1.47\times 10^3}{3\times 10^8}=\bf 4.90\;\rm \mu s$$
This shows that the fraction of muons reaching the ground is
$$\left[0.5\right]^{t'/T}=\left[0.5\right]^{4.9\mu/1.5\mu}=0.10$$
which is the same known fraction.
