Let the infinite geometric series be $a + ar + ar^2 + \cdots$ where $a>0$ and $0<r<1$ since all terms are positive and the infinite sum exists.

Use the given sum of the geometric series $\frac{a}{1-r}=3$ So, $a=3(1-r).$

Form the series of cubes of the terms The cubes are: $a^3,\ a^3r^3,\ a^3r^6,\dots$ This is also an infinite geometric series with first term $a^3$ and common ratio $r^3$.

Its sum is given as $\frac{a^3}{1-r^3}=\frac{27}{19}.$

Substitute $a=3(1-r)$ $\frac{[3(1-r)]^3}{1-r^3}=\frac{27}{19}$ $\frac{27(1-r)^3}{1-r^3}=\frac{27}{19}$

Cancel $27$ from both sides: $\frac{(1-r)^3}{1-r^3}=\frac{1}{19}.$

Factorize $1-r^3$ $1-r^3=(1-r)(1+r+r^2).$

Hence, $\frac{(1-r)^3}{(1-r)(1+r+r^2)}=\frac{1}{19}$ $\frac{(1-r)^2}{1+r+r^2}=\frac{1}{19}.$

Cross-multiply and simplify $19(1-r)^2=1+r+r^2$ $19(1-2r+r^2)=1+r+r^2$ $19-38r+19r^2=1+r+r^2$ $18-39r+18r^2=0$ Divide by $3$: $6-13r+6r^2=0$ $6r^2-13r+6=0.$

Solve the quadratic $6r^2-13r+6=(3r-2)(2r-3)=0$ So, $r=\frac{2}{3} \quad \text{or} \quad r=\frac{3}{2}.$

Since for an infinite geometric series with positive terms we need $0<r<1$, only $r=\frac{2}{3}$ is valid.

Check with options The correct option is: $\boxed{\frac{2}{3}}$ which is Option C.