Answer
$\beta = 0.750$
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 $\Delta t'$:
$\Delta t' = t_B'-t_A'$
$\Delta t' = \gamma~[(t_B-t_A)-\beta~(x_B-x_A)/c]$
$\Delta t' = \gamma~[(1.00~\mu s)-\beta~(400~m)/c]$
We can find $\beta$ when $\Delta t' = 0$:
$\Delta t' = \gamma~[(1.00~\mu s)-\beta~(400~m)/c] = 0$
$(1.00~\mu s)-\beta~(400~m)/c = 0$
$(1.00~\mu s) = \beta~(400~m)/c$
$\beta = \frac{(3.0\times 10^8~m/s)~(1.00\times 10^{-6}~s)}{400~m}$
$\beta = 0.750$
