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

replace $x$ by $\frac{2}{x}$ in above identity :-

$\frac{2^{10}\left(2 x^{2}+3 x+4\right)^{10}}{x^{20}}=\sum_{r=0}^{20} \frac{a_{r} 2^{r}}{x^{r}}$

$\Rightarrow 2^{10} \sum_{\mathrm{r}=0}^{20} \mathrm{a}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}=\sum_{\mathrm{r}=0}^{20} \mathrm{a}_{\mathrm{r}} 2^{\mathrm{r}} \mathrm{x}^{(20-\mathrm{r})}$ (from (i))

now, comparing coefficient of $x^{7}$ from both sides

(take $r=7$ in L.H.S. \& $r=13$ in R.H.S.)

$2^{10} \mathrm{a}_{7}=\mathrm{a}_{13} 2^{13} \Rightarrow \frac{\mathrm{a}_{7}}{\mathrm{a}_{13}}=2^{3}=8$