the minimum cable bandwidth
First we can calculate how many bits we need to represent a single sample with maximum 0.25% error: 100/0.25=400, so we need 400 different quantization values,
to express them in a binary-coded form log_2(400) ~= 8.64 bits are necessary.

If we assume no compression is used (none is mentioned in the text) this means 9 bits. We need to transmit 4*20000 samples per second per mic (the 4 comes from "twice the Nyquist rate" the Nyquist rate is twice the signal bandiwdth of 20kHz). Since there are 11 time-dicision multiplexed mics this means 11*4*20kHz*9bits = 7920000 bits/sec = 7.92 MB/s.