Answer
Choice D.
Work Step by Step
Assuming the annihilation takes place on the line, one photon travels 6 cm longer than the other one.
This corresponds to a later arrival time of $\Delta t = \frac{0.06 m}{c}$. Plugging in $c=3.00\times10^8 m/s$, we find a time of 0.2 ns. This is less than 10 ns, so the two photons will be counted as simultaneous. This is choice D.
