Answer
$\Delta t_0 =0.446ps$
Work Step by Step
First we need to get the time from the perspective of the particle detector with the given velocity and distance.
$\Delta t_0 = 0.00105m/(0.0092c)$
Using a the time dilation equation given we can solve for the "proper time":
$\Delta t=\Delta t_0 / \sqrt{1-(v/c)^2 }$
$\sqrt{1-(v/c)^2 } \Delta t=\Delta t_0 $
$\Delta t_0 =\sqrt{1-(v/c)^2 } \Delta t$
Now we substitute in our known values:
$\Delta t_0 =\sqrt{1-(.992c/c)^2 } \frac{0.00105m}{.992c})$
$\Delta t_0 =4.46*10^{-13}s$
$\Delta t_0 =0.446ps$
