The disk in question has a transfer rate of 80 MiB/s, or 81920 kiB/s, or 20480 blocks/s. Here, we will round off to 20,000 blocks/second, since this appears to be what your professor did. This equates to 0.05 ms to transfer a block, explaining the last term in the equation.
Finally, in addition to seek time (time to move the drive head to the track), there is also the rotational latency of the disk itself to deal with. At 7200 RPM, in the worst case, we have to wait 1 full revolution, but on average we have to wait a half-revolution - or 4.166ms (7200 RPM = 120 rev/sec = 8.333ms/rev).
Thus, to transfer one block to the computer, we must wait the equivalent
Seek Time +
Rotational Latency +
5 ms + 4.166 ms + 0.05 ms = 9.216 ms
Note that for solid-state drives, while there is no rotational latency to take into account, there certainly still is a measurable seek time (to actually address the contents of the sectors in the flash memory) and transfer time (largely limited by the bus being used to transfer the data itself, e.g. SATA).
Thus, in general, the total access time to read a single sector for a drive is (neglecting software):
Rotational/Hard Drive: Seek Time + Rotational Latency + Transfer Time
Solid-State Drive: Seek Time + Transfer Time