Answer
The spatial coordinate of the event according to S' is $~~x' = 138~km$
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 spatial coordinate $x'$:
$x' = \gamma~(x-vt)$
$x' = 3.20~[100~km-(0.950)(3.0\times 10^8~m/s)(200\times 10^{-6}~s)]$
$x' = (3.20)(43~km)$
$x' = 138~km$
The spatial coordinate of the event according to S' is $~~x' = 138~km$
