Answer
$1.97ns$
Work Step by Step
We know that
$\Delta t-\Delta t_{\circ}=\frac{v^2}{2c^2}\Delta t_{\circ}$
We plug in the known values to obtain:
$\Delta t-\Delta t_{\circ}=\frac{(222m/s)^2}{2(3\times 10^8m/s)^2}(2hr)$
$\Delta t-\Delta t_{\circ}=(5.476\times 10^{-13}hr)(3600s/hr)$
$\Delta t-\Delta t_{\circ}=1.97136\times 10^{-9}s=1.97ns$
