Answer
$t' = -3.1\times 10^{-6}~s$
Work Step by Step
We can find $\gamma$:
$\gamma = \frac{1}{1-\beta^2}$
$\gamma = \frac{1}{1-0.60^2}$
$\gamma = 1.5625$
We can find the temporal coordinate $t'$ of event 2:
$t' = \gamma~(t-vx/c^2)$
$t' = 1.5625~[(4.0\times 10^{-6}~s)-\frac{(0.60)(3.0\times 10^8~m/s)(3000~m)}{(3.0\times 10^8~m/s)^2}]$
$t' = (1.5625)(-2.0\times 10^{-6}~s)$
$t' = -3.1\times 10^{-6}~s$
