Hint: don't use multinomial

We have $(2x^{2} + 3x + 4)^{10} = \sum_{r=0}^{20} a_{r}x^{r}$ ...(1)

Replace $x$ by $\frac{2}{x}$ , we get

$(\frac{8}{x^{2}} + \frac{6}{x} + 4)^{10} = \sum_{r=0}^{20} a_{r}(\frac{2}{x})^{r}$

$2^{10}(2x^{2} + 3x + 4)^{10} = \sum_{r=0}^{20} a_{r}.2^{r}.x^{20-r}$

Using equation (1)

$2^{10}\sum_{r=0}^{20} a_{r}.x^{r} = \sum_{r=0}^{20} a_{r}.2^{r}.x^{20-r}$

Comparing the coefficient of $x^{7}$ in both sides

$2^{10}a_{7} = 2^{13}a_{13}$

$a_{7} = a_{13}2^{3}$ $\Rightarrow$ $\frac{a_{7}}{a_{13}} = \boxed{8}$ .