Correct Answer - Option 1 : 20 Mbps

__Concept__:

The capacity of a band-limited AWGN channel is given by the formula:

\(C = B{\log _2}\left( {1 + \frac{S}{N}} \right)\)

C = Maximum achievable data rate (in bits/sec)

B = channel bandwidth

\(\frac{S}{N}\) = Signal to Noise power

Note: In the expression of channel capacity, S/N is not in dB.

__Calculation__:

Given B.W. = 4 MHz, S/N = 31

The channel capacity will be:

\(C = (4\times 10^6)~{\log _2}\left( {1 + 31} \right)\)

**C = 20 Mbps**