Answer
The arrival of the optical signal precedes the arrival of the train by $2.2\times 10^{-6}~s$
Work Step by Step
We can find $T_L$, the time it takes the light signal to reach the station according to the stationmaster's clock:
$T_L = \frac{10^3~m}{3.0\times 10^8~m/s}$
$T_L = 3.3\times 10^{-6}~s$
We can find $T_T$, the time it takes the train to reach the station according to the stationmaster's clock:
$T_T = \frac{10^3~m}{(0.6)(3.0\times 10^8~m/s)}$
$T_T = 5.5\times 10^{-6}~s$
The time difference is $T_T-T_L$ which is $2.2\times 10^{-6}~s$
The arrival of the optical signal precedes the arrival of the train by $2.2\times 10^{-6}~s$.
