The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $$\frac{27}{19}$$. Then the common ratio of this series is:

We have an infinite geometric progression whose first term is $$a$$ and common ratio is $$r$$, with $$0<r<1$$ because all terms are positive and the series converges.

For an infinite GP, the standard formula for the sum is

$$S_\infty=\dfrac{a}{1-r}.$$

Here it is given that this sum equals $$3$$. So

$$\dfrac{a}{1-r}=3 \quad\Longrightarrow\quad a=3(1-r).$$

Next, we are told that the sum of the cubes of the terms is $$\dfrac{27}{19}$$. When each term of a GP is cubed, we again obtain a GP, now with first term $$a^3$$ and common ratio $$r^3$$. The sum of this new GP is therefore

$$S_\infty^{(\text{cubes})}=\dfrac{a^3}{1-r^3}.$$

Using the given value, we write

$$\dfrac{a^3}{1-r^3}=\dfrac{27}{19}.$$

Substituting $$a=3(1-r)$$ into this equation gives

$$\dfrac{\bigl(3(1-r)\bigr)^3}{1-r^3}=\dfrac{27}{19}.$$

Simplifying the numerator,

$$\dfrac{27(1-r)^3}{1-r^3}=\dfrac{27}{19}.$$

We can cancel the common factor $$27$$ on both sides:

$$\dfrac{(1-r)^3}{1-r^3}=\dfrac{1}{19}.$$

Now, recall the algebraic identity

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

Substituting this into the denominator, we obtain

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

One factor of $$1-r$$ cancels, leaving

$$\dfrac{(1-r)^2}{1+r+r^2}=\dfrac{1}{19}.$$

Cross-multiplying gives

$$19(1-r)^2 = 1 + r + r^2.$$

Expanding the square in the left-hand side,

$$19\bigl(1 - 2r + r^2\bigr) = 1 + r + r^2.$$

Distributing the $$19$$, we have

$$19 - 38r + 19r^2 = 1 + r + r^2.$$

Bringing all terms to one side,

$$19 - 38r + 19r^2 - 1 - r - r^2 = 0.$$

Combining like terms,

$$18 - 39r + 18r^2 = 0.$$

To simplify, divide every term by $$3$$:

$$6 - 13r + 6r^2 = 0.$$

This is a quadratic in $$r$$. Using the quadratic formula $$r=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$ with $$a=6,\; b=-13,\; c=6,$$ we get

$$r=\dfrac{13 \pm \sqrt{(-13)^2 - 4\cdot6\cdot6}}{2\cdot6} =\dfrac{13 \pm \sqrt{169 - 144}}{12} =\dfrac{13 \pm \sqrt{25}}{12} =\dfrac{13 \pm 5}{12}.$$

This yields two possible values:

$$r_1=\dfrac{13+5}{12}=\dfrac{18}{12}=\dfrac{3}{2},\qquad r_2=\dfrac{13-5}{12}=\dfrac{8}{12}=\dfrac{2}{3}.$$

Since the common ratio of a convergent infinite geometric series must satisfy $$0<r<1$$, we discard $$\dfrac{3}{2}$$ and keep

$$r=\dfrac{2}{3}.$$

Hence, the correct answer is Option B.