To go to higher and higher speeds, it is not possible to just keep increasing the sampling rate. The Nyquist theorem says that even with a perfect 3000-Hz line (which a dial-up telephone is decidedly not), there is no point in sampling faster than 6000 Hz. In practice, most modems sample 2400 times/sec and focus on getting more bits per sample.
The number of samples per second is measured in baud. During each baud, one symbol is sent. Thus, an n-baud line transmits n symbols/sec. For example, a 2400-baud line sends one symbol about every 416.667 µsec. If the symbol consists of 0 volts for a logical 0 and 1 volt for a logical 1, the bit rate is 2400 bps. If, however, the voltages 0, 1, 2, and 3 volts are used, every symbol consists of 2 bits, so a 2400-baud line can transmit 2400 symbols/sec at a data rate of 4800 bps. Similarly, with four possible phase shifts, there are also 2 bits/symbol, so again here the bit rate is twice the baud rate. The latter technique is widely used and called QPSK (Quadrature Phase Shift Keying).
-- Tannenbaum
Portanto, temos uma linha de 6MHz. O teorema de Nyquist diz que não faz sentido usar uma amostragem maior que 12.000.000 Hz (12 milhões de amostragens por segundo) ou 12Mbps (bauds/seg).
O sinal digital da questão tem 4 níveis (2 bits) que nos dá uma taxa teórica de envio de 2*12Mbps = 24Mbps