Let $\alpha ={\sin }^{-1}\left(\frac{3}{5}\right)$ and $\beta =2{\cos }^{-1}\left(\frac{2}{\sqrt{5}}\right)$, so $\cos \frac{\beta }{2}=\frac{2}{\sqrt{5}}$.

From $\sin \alpha =\frac{3}{5}$ we get $\tan \alpha =\frac{3}{4}$.

For $\beta$: with $\cos \frac{\beta }{2}=\frac{2}{\sqrt{5}}$ and $\sin \frac{\beta }{2}=\frac{1}{\sqrt{5}}$, so $\tan \frac{\beta }{2}=\frac{1}{2}$. Using the double-angle identity,

$\tan \beta =\frac{2\tan (\beta /2)}{1-{\tan }^2(\beta /2)}=\frac{2\cdot \frac{1}{2}}{1-\frac{1}{4}}=\frac{4}{3}.$

Finally,

$\tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }=\frac{\frac{3}{4}-\frac{4}{3}}{1+\frac{3}{4}\cdot \frac{4}{3}}=\frac{-7/12}{2}=-\frac{7}{24}.$

$\therefore$ The answer is option (B).