Answer
$t' = 0$
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 1:
$t' = \gamma~(t-vx/c^2)$
$t' = 1.5625~[(0)-\frac{(0.60)(3.0\times 10^8~m/s)(0)}{(3.0\times 10^8~m/s)^2}]$
$t' = 0$
