We know from Shannon's Theorem,Maximum data rate of a channel in bps (B) = Hlog2 ( 1+S/N ) _ 1Where H = bandwidth in HzS/N = signal-to-noise ratioWe also know thatDb = 10log10 S/NWhere Db= signal-to-noise ratio in decibel (which is in this case 20)So20 = 10 log10 S/N2 = log10 S/NAnd we get S/N = 100Putting this value of S/N in equation 1 we getB = 6000 log2 (1+100) = 6000 x 6.6582= 39949.2 bps or 39.949 kbpsSo the maximum achievable data rate here is 39.949 kbps