We are asked to find the ratio of the coefficient of the $x^5$ term to the coefficient of the $x^3$ term in the binomial expansion of $(2x + 3)^6$.

1. Problem Description

We are asked to find the ratio of the coefficient of the $x^5$ term to the coefficient of the $x^3$ term in the binomial expansion of $(2x + 3)^6$.

2. Solution Steps

The binomial expansion of $(a+b)^n$ is given by:

$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

In our case, $a = 2x$, $b = 3$, and $n = 6$.

To find the term with $x^5$, we need $n-k = 5$, so $6-k = 5$, which means $k = 1$.

The $x^5$ term is:

$\binom{6}{1} (2x)^{6-1} (3)^1 = \binom{6}{1} (2x)^5 (3)^1 = 6 \cdot 2^5 x^5 \cdot 3 = 6 \cdot 32 \cdot 3 x^5 = 576x^5$

To find the term with $x^3$, we need $n-k = 3$, so $6-k = 3$, which means $k = 3$.

The $x^3$ term is:

$\binom{6}{3} (2x)^{6-3} (3)^3 = \binom{6}{3} (2x)^3 (3)^3 = \frac{6!}{3!3!} (8x^3) (27) = \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} (8x^3) (27) = 20 \cdot 8 \cdot 27 x^3 = 4320x^3$

The ratio of the term in $x^5$ to the term in $x^3$ is:

$\frac{576x^5}{4320x^3} = \frac{576}{4320} x^{5-3} = \frac{576}{4320} x^2 = \frac{1}{7.5} x^2 = \frac{2}{15} x^2$

However, the question asks for the ratio of the coefficient of $x^5$ to the coefficient of $x^3$.

So we need to find $\frac{576}{4320}$.

Simplifying this fraction, we get $\frac{576}{4320} = \frac{288}{2160} = \frac{144}{1080} = \frac{72}{540} = \frac{36}{270} = \frac{18}{135} = \frac{6}{45} = \frac{2}{15}$.

Thus, the ratio of the coefficients is $\frac{2}{15}$. The given options are $\frac{x^2}{3}, 2x^2, \frac{x^3}{5}, 2$, none of which match $\frac{2}{15}x^2$. Also, the options don't include $\frac{2}{15}$.

If the question asked for the ratio of the coefficient of $x^5$ to the coefficient of $x^3$, which is $\frac{576}{4320} = \frac{2}{15}$.

Let us check the calculation. The general term is $\binom{6}{r} (2x)^{6-r} 3^r = \binom{6}{r} 2^{6-r} 3^r x^{6-r}$.

If $x^5$, then $6-r=5$ so $r=1$. The coefficient is $\binom{6}{1} 2^{6-1} 3^1 = 6 \cdot 2^5 \cdot 3 = 6 \cdot 32 \cdot 3 = 576$.

If $x^3$, then $6-r=3$ so $r=3$. The coefficient is $\binom{6}{3} 2^{6-3} 3^3 = 20 \cdot 2^3 \cdot 3^3 = 20 \cdot 8 \cdot 27 = 4320$.

The ratio is indeed $\frac{576}{4320} = \frac{2}{15}$.

If the ratio is for terms, we have $\frac{576 x^5}{4320 x^3} = \frac{2}{15} x^2$. If $\frac{x^2}{a} = \frac{2 x^2}{15}$, then $a = \frac{15}{2} = 7.5$. But this is still not in the options.

3. Final Answer

The ratio of the coefficient of $x^5$ to the coefficient of $x^3$ is $\frac{2}{15}$. However, the options provided are $\frac{x^2}{3}, 2x^2, \frac{x^3}{5}, 2$, and the ratio of the terms is $\frac{2}{15} x^2$. Thus, the question or the options are likely incorrect. Assuming the question intended to ask for the ratio of the coefficients, there is no answer among the choices.