Answer
$\Delta t' = \gamma~[(1.00~\mu s)-\beta~(400~m)/c]$
Work Step by Step
We can write an expression for $t_A'$:
$t_A' = \gamma~[t_A-c\beta~x_A/c^2] = \gamma~[t_A-\beta~x_A/c]$
We can write an expression for $t_B'$:
$t_B' = \gamma~[t_B-c\beta~x_B/c^2] = \gamma~[t_B-\beta~x_B/c]$
We can write an expression for $\Delta t'$:
$\Delta t' = t_B'-t_A'$
$\Delta t' = \gamma~[(t_B-t_A)-\beta~(x_B-x_A)/c]$
$\Delta t' = \gamma~[(1.00~\mu s)-\beta~(400~m)/c]$
