Find Bandwith from Spectral Efficiency and Eb/N0
Your initial answer is correct -- the bandwidth depends only on the spectral efficiency and the symbol rate. $E_b/N_0$ is a figure of merit largely designed to be independent of the bandwidth.

Let me give you an example of where $E_b/N_0$ might come into play. Say you are required to provide $E_b/N_0 = 10$, and you know that your receiver is rated at $N_0 = 10 ^ {-12}$. Then, $E_b = 10 N_0 = 10^{-11}$. At a rate $R = 2 \times 10^6$, the received power is $(2 \times 10^6)(10^{-11} = 2 \times 10^{-5}$ watts, or $-47$ dB.

If you know the channel attenuation, then you can calculate the transmitted power. From there, you can calculate things like battery life. And so on...