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