Seems that the table might be related to additional information provided by the author of the book. Unfortunately I don't have this book, therefore I'll just try to describe what do I understand from the table.
What do we know about the channel? We know that its upper cut-off frequency is 3kHz. Also (due to absence of information regarding the lower cut-off frequency) we can assume that the channel can transmit down to DC (0Hz).
"T" column
Now, we want to see how much time will it take to transmit 8 bits (1 byte) of data at different bit-rates. The equation is very simple:
$$T_{byte}[sec]=\frac{1}{Bps}*8$$
There should be nothing confusing about this column - the higher the bit rate, the shorter the transmission time.
"First harmonic" column
We also want to know what is the lowest frequency associated with transmitting information in 8 bits chunks over the channel. The formula is also pretty straightforward in this case:
$$f_{lowest}[Hz]=\frac {1}{T_{byte}}*\#cycles\_in\_one\_byte$$
Why is this frequency depends on the # of cycles? Compare the "switching pattern" corresponding to the byte 10101010 with this corresponding to byte 11001100. It is evident that there are more transitions of the signal in the first byte. One cycle of a wave is usually defined as a single 10 pattern: the first byte requires 4 full cycles in order to be transmitted, while the second byte requires just two (but the period of the wave is twice longer). According to the formula above, since the first byte requires twice more cycles to be transmitted, the frequency corresponding to this byte is twice higher.
Assuming that the lowest frequency corresponds to byte 11110000 (which can be though of as a single cycle signal 10 having longer period), we can see that there is just a single cycle in this pattern, thus the formula becomes:
$$f_{lowest}[Hz]=\frac {1}{T_{byte}}$$
"# Harmonics sent" column
This is the most confusing column in the table, which requires the whole context of the discussion in the book in order to be interpreted.
My guess is that the author wanted to show that if you make use of just the fraction of available BW, then you still can use the remaining BW for data transmission. The technique of partitioning a single physical channel into multiple transmission channels is called Frequency-division multiplexing.
In case of FDM, the total bit-rate becomes:
$$Bps_{total}=Bps_{channel}*\#Channels$$
For example: if you communicate at 300 Bps, then (theoretically, according to the author) you could allocate up to 80 transmission channels inside the initial BW. This would give rise to a total bit-rate of \$Bps_{total}(300)=300*80=24000\frac{bits}{sec}\$.
If, on the other hand, you transmit at 19200 Bps, then you won't be able to allocate the second channel having the same bit-rate into the given BW, therefore getting a total bit-rate of \$Bps_{total}(19200)=19200\frac{bits}{sec}\$.
You can see that the total attainable bit-rates have the same order of magnitude, although the bit-rates for a transfer of the actual information differ significantly. This is not a coincidence - the maximal bit-rate of a physical channel depends on the BW, which is the same in both cases (and you can check for other bit rates in the table that the total bit-rate will be approximately the same). The differences in the total bit-rates for different rows in the table arise due to the fact that there will be different unused portions of the BW (the remainder of dividing 3kHz by the frequency of the first harmonic).
In summary
Assuming that the author of the book did explain the idea behind FDM at some earlier stage in the book, this table makes perfect sense and shows that BW indeed provides a measure of the attainable data rates on the channel.
The actual bit-rate at which the data is transmitted is not that important, because you can always use the remaining BW and allocate additional transmission channels in it.
Note: the discussion here is purely theoretical. The actual implementation of FDM schemes can be overwhelmingly complicated and reduce the initial BW (by introducing guard bands for example).