Answer
The light from bolt 1 reaches the assistant $4.0\times 10^{-5}~s$ earlier.
Work Step by Step
We can find the time it takes the light from bolt 1 to reach you:
$t_1 = \frac{d}{c} = \frac{9000~m}{3.0\times 10^8~m/s} = 3.0\times 10^{-5}~s$
We can find the time it takes the light from bolt 2 to reach you:
$t_2 = \frac{d}{c} = \frac{3000~m}{3.0\times 10^8~m/s} = 1.0\times 10^{-5}~s$
Since the light arrives at the same time, the flash from bolt 1 must have left $2.0\times 10^{-5}~s$ earlier.
We can find the time it takes the light from bolt 1 to reach the assistant:
$t_1 = \frac{d}{c} = \frac{3000~m}{3.0\times 10^8~m/s} = 1.0\times 10^{-5}~s$
We can find the time it takes the light from bolt 2 to reach the assistant:
$t_2 = \frac{d}{c} = \frac{9000~m}{3.0\times 10^8~m/s} = 3.0\times 10^{-5}~s$
Since the light from bolt 1 left $2.0\times 10^{-5}~s$ earlier, and it takes $2.0\times 10^{-5}~s$ less time to travel to the assistant, the light from bolt 1 reaches the assistant $4.0\times 10^{-5}~s$ earlier.
