Answer
The temporal coordinate of the event according to S' is $~~t' = -3.73\times 10^{-4}~s$
Work Step by Step
We can find $\gamma$:
$\gamma = \frac{1}{1-\beta^2}$
$\gamma = \frac{1}{1-0.950^2}$
$\gamma = 3.20$
We can find the temporal coordinate $t'$:
$t' = \gamma~(t-vx/c^2)$
$t' = 3.20~[(200\times 10^{-6}~s)-\frac{(0.950)(3.0\times 10^8~m/s)(1.00\times 10^5~m)}{(3.0\times 10^8~m/s)^2}]$
$t' = (3.20)(-1.167\times 10^{-4}~s)$
$t' = -3.73\times 10^{-4}~s$
The temporal coordinate of the event according to S' is $~~t' = -3.73\times 10^{-4}~s$
