As mentioned above, you should make a distinction between the legal limits, and the technical feasibility, and the theoretical maxima in rates.
To illustrate: Shannon's channel capacity says that over a channel with bandwidth $B$ and SNR $\frac SN$, you can do at most the bit rate $C$:
$$C = B\log_2\left(1+\frac SN\right)\,\text.$$
What Shannon does not say is how you do that, and in fact, that problem is often unsolved, so most transceiver systems stay considerably below that rate limit.
Let us, however, plug in some numbers here of a rather idyllic scenario:
We go for a channel width of $B= 3\,\text{kHz} \approx 35\,\text{dBHz}$, and let's assume some ~S8 scenario, so $S=-80\,\text{dBm}$, as well as letting your receiver run at room temperature and have a Noise Figure $N_F=4\,\text{dB}$:
$$\begin{align}
C &= 3\cdot10^3\frac1{\text s} \cdot\log_2\left(1+\frac{-80\,\text{dBm}}{-174\,\text{dBm/Hz}\cdot 35\,\text{dBHz}\cdot4\,\text{dB}}\right)\\
&= 3\cdot10^3\frac1{\text s} \cdot\log_2\left(1+\frac{-80\,\text{dBm}}{\left(-174+35+4\right)\,\text{dBm}}\right)\\
&= 3\cdot10^3\frac1{\text s} \cdot\log_2\left(1+(-80-(-174+35+4)\,\text{dB})\right)\\
&= 3\cdot10^3\frac1{\text s} \cdot\log_2\left(1+(-80+135)\,\text{dB})\right)\\
&= 3\cdot10^3\frac1{\text s} \cdot\log_2\left(1+55\,\text{dB}\right)\\
&\approx 3\cdot10^3\frac1{\text s} \cdot\log_2\left(10^{5.5}\right)\\
&\approx 3\cdot10^3\frac1{\text s} \cdot18.27\\
&= 54.81\,\text{kb/s}
\end{align}$$
In other words, no matter what modulation, channel code you use, there's no chance you'll get more that 54810 bits per second error-free through that channel. So, restricting you to about $\frac13$ of that still makes no sense from an RF perspective (regulator doesn't actually care how fast you transmit, but how much spectrum you occupy with that), but it's not inherently too optimistic – on a short range scenario, the receive powers might be much higher (calculate Free Space Path loss for 5 km with 2 m waves, how much power do you get when someone blasts out 750 W? My rough head calculation says you get about 20 times the capacity).