Capacity computation in Costa sheme
You really didn't give sufficient context and background information to provide a complete answer, but I'll try anyway. You didn't say so, but I assume that $\sigma_W^2$ is the power of a watermark and that $\sigma_N^2$ is the power of additive noise. I furthermore assume that the "usable bandwidth" is the bandwidth for generating the watermark. If we additionally assume that the watermark signal is white with spectral density $W_0$, then its power inside the usable bandwidth $B$ is given by

$$\sigma_W^2=2W_0B$$

If $B_1=22050Hz$ and $B_2=4000Hz$ we get for the capacity ratio

$$\frac{\log_2(1+\frac{2W_0B_2}{\sigma_N^2})}{\log_2(1+\frac{2W_0B_1}{\sigma_N^2})}$$

which is not a fixed ratio, but it depends on the constants $W_0$ and $\sigma_N^2$. If they are not given, then there is no way to claim that this ratio is $1/5$.

However, if you can assume that $\sigma_W^2/\sigma_N^2\ll 1$ then the capacity can be approximated by

$$\frac{1}{2\log (2)}\frac{\sigma_W^2}{\sigma_N^2}=\frac{1}{2\log (2)}\frac{2W_0B}{\sigma_N^2}$$

and with the same assumptions as above the capacity ratio is equal to the ratio of bandwidths $B_2/B_1=0.18$, which is indeed close to $1/5$.

If any of my assumptions are not correct, please clarify your question.