Answer
$1.02\times10^{-4}m$.
Work Step by Step
First, from the KE, calculate the speed of the tau lepton. Use equation 26–5a and solve for v.
$$v=c\sqrt{1-\frac{1}{\left( \frac{KE}{mc^2}+1\right)^2}}$$
$$v=c\sqrt{1-\frac{1}{\left( \frac{950MeV}{1777MeV}+1\right)^2}}=0.7585c$$
Find the dilated lifetime of the particle in the lab. It will be longer than the proper time.
$$\Delta t_{lab}=\frac{\Delta t_0}{\sqrt{1-v^2/c^2}}$$
$$=\frac{2.91\times10^{-13}s}{\sqrt{1-0.7585^2}}=4.465\times10^{-13}s$$
Now use the speed and time together to fin the length of the particle track.
$$\Delta x_{lab}=v(\Delta t_{lab})=(0.7585)(3.00\times10^8m/s)( 4.465\times10^{-13}s)$$
$$=1.02\times10^{-4}m$$
