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

1. Problem Description

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

2. Solution Steps

The general term in the binomial expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.

For the term with $x^5$ in the expansion of $(2x+3)^6$, we need $n-r = 5$, so $6-r=5$, which means $r=1$.

So the term with $x^5$ is $T_{1+1} = T_2 = \binom{6}{1} (2x)^5 (3)^1 = 6 \cdot 32x^5 \cdot 3 = 576x^5$.

For the term with $x^3$ in the expansion of $(2x+3)^6$, we need $n-r = 3$, so $6-r=3$, which means $r=3$.

So the term with $x^3$ is $T_{3+1} = T_4 = \binom{6}{3} (2x)^3 (3)^3 = \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} \cdot 8x^3 \cdot 27 = 20 \cdot 8x^3 \cdot 27 = 4320x^3$.

The ratio of the term in $x^5$ to the term in $x^3$ is $\frac{576x^5}{4320x^3}$. However, the problem asks for the *ratio of the terms*, which implies that we are interested in comparing the coefficients, so we compute $\frac{576}{4320}$.

$\frac{576}{4320} = \frac{576 \div 144}{4320 \div 144} = \frac{4}{30} = \frac{2}{15}$.

Thus the ratio is $\frac{2x^2}{15}$.

3. Final Answer

B. $\frac{2x^2}{15}$