We have been given that ${}^{2n}C_3:{}^{n}C_3=10$

We know that ${}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}$

$\Rightarrow \frac{{\displaystyle \frac{\left(2n\right)!}{3!\left(2n-3\right)!}}}{{\displaystyle \frac{n!}{3!\left(n-3\right)!}}}=10$

$\Rightarrow \frac{\left(2n\right)!}{3!\left(2n-3\right)!}\times \frac{3!\left(n-3\right)!}{n!}=10$

$\Rightarrow \frac{\left(2n\right)\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)!}{\left(2n-3\right)!}\times \frac{\left(n-3\right)!}{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}=10$

$\Rightarrow \frac{\left(2n\right)\left(2n-1\right)\left(2n-2\right)}{n\left(n-1\right)\left(n-2\right)}=10$

$\Rightarrow \frac{\left(4\right)\left(2n-1\right)}{\left(n-2\right)}=10$

$\Rightarrow 8n-4=10n-20$

$\Rightarrow n=8$

Hence, the value of $\frac{n^2+3n}{n^2-3n+4}$ is $=\frac{8^2+3\left(8\right)}{8^2-3\left(8\right)+4}=\frac{88}{44}=2$

Therefore, the value of $\frac{n^2+3n}{n^2-3n+4}$ is $2:1$.