The ratio of the coefficient of $$x^{15}$$ to the term independent of $$x$$ in the expansion of $$\left(x^2 + \frac{2}{x}\right)^{15}$$ is:

We are given the binomial expansion of $$\left(x^2 + \frac{2}{x}\right)^{15}$$. The general term in the expansion is given by the binomial theorem. For $$(a + b)^n$$, the $$(r+1)^{\text{th}}$$ term is $$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$. Here, $$a = x^2$$, $$b = \frac{2}{x}$$, and $$n = 15$$.

So, the general term is:

$$T_{r+1} = \binom{15}{r} (x^2)^{15-r} \left(\frac{2}{x}\right)^r$$

Simplify the exponents:

$$T_{r+1} = \binom{15}{r} x^{2(15-r)} \cdot \frac{2^r}{x^r} = \binom{15}{r} x^{30 - 2r} \cdot 2^r \cdot x^{-r} = \binom{15}{r} 2^r x^{30 - 3r}$$

The exponent of $$x$$ is $$30 - 3r$$.

First, we find the coefficient of $$x^{15}$$. Set the exponent equal to 15:

$$30 - 3r = 15$$

Solve for $$r$$:

$$30 - 15 = 3r \implies 15 = 3r \implies r = 5$$

Substitute $$r = 5$$ into the general term to get the coefficient:

$$\text{Coefficient of } x^{15} = \binom{15}{5} 2^5$$

Next, we find the term independent of $$x$$, which means the exponent of $$x$$ is 0. Set the exponent equal to 0:

$$30 - 3r = 0$$

Solve for $$r$$:

$$30 = 3r \implies r = 10$$

Substitute $$r = 10$$ into the general term to get the coefficient:

$$\text{Coefficient of constant term} = \binom{15}{10} 2^{10}$$

Note that $$\binom{15}{10} = \binom{15}{5}$$ because $$\binom{n}{r} = \binom{n}{n-r}$$. So, $$\binom{15}{10} = \binom{15}{5}$$.

The ratio required is the coefficient of $$x^{15}$$ to the coefficient of the constant term:

$$\text{Ratio} = \frac{\binom{15}{5} \cdot 2^5}{\binom{15}{10} \cdot 2^{10}} = \frac{\binom{15}{5} \cdot 32}{\binom{15}{5} \cdot 1024}$$

Since $$\binom{15}{5}$$ is common and non-zero, it cancels out:

$$\text{Ratio} = \frac{32}{1024}$$

Simplify the fraction by dividing both numerator and denominator by 32:

$$\frac{32 \div 32}{1024 \div 32} = \frac{1}{32}$$

Thus, the ratio is $$1 : 32$$.

Comparing with the options, this matches Option D.

Hence, the correct answer is Option D.