Answer
The time that passes in the cosmic ray's reference frame is $~346~\mu s$
Work Step by Step
We can find the speed of the cosmic ray according to the Earth's reference frame:
$v = \frac{d}{t} = \frac{60\times 10^3~m}{400\times 10^{-6}~s} = 1.5\times 10^8~m/s = 0.5~c$
We can find the time $t_0$ that passes in the cosmic ray's reference frame:
$t' = \gamma~t_0$
$t_0 = \frac{t'}{\gamma}$
$t_0 = t'~\sqrt{1-\frac{v^2}{c^2}}$
$t_0 = t'~\sqrt{1-\frac{(0.5~c)^2}{c^2}}$
$t_0 = (400~\mu s)~\sqrt{1-(0.5)^2}$
$t_0 = 346~\mu s$
The time that passes in the cosmic ray's reference frame is $~346~\mu s$
