Answer
$2.07\times10^{-6}s$
Work Step by Step
You measure the dilated time interval. Find the proper time interval using equation 26–1a.
$$\Delta t_o = \Delta t \sqrt{1-\frac{ v^{2} } { c^{2} }} $$
$$= (4.76\times10^{-6}s) \sqrt{1-\frac{ (2.70\times10^8m/s)^{2} } { (3.00\times10^8m/s)^{2} }} $$
$$= 2.07\times10^{-6}s$$