To determine the theoretical highest data rate for a noisy channel we use the Shannon capacity formula.
$\text{Capacity=BW*$\log_2$(1+SNR)}$
where; 1) BW=Bandwidth of the channel
2) SNR= signal to noise ratio
3) Capacity= Capacity of the channel in bit per second
$\therefore \text{Capacity=6*$\log_2$(1+16)}$
$\text{Capacity=6*$\log_2$(17) $\approx$24.74 kbps}$
So option $D$ is correct here.
