The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of

$\frac{a}{1-r}=3$           ........(1)

$\frac{\mathrm{a}^{3}}{1-\mathrm{r}^{3}}=\frac{27}{19} \Rightarrow \frac{27(1-\mathrm{r})^{3}}{1-\mathrm{r}^{3}}=\frac{27}{19}$

$\Rightarrow 6 r^{2}-13 r+6=0$

$\Rightarrow r=\frac{2}{3}$ as $|r|<1$