Given,

The circles ${\mathrm{C}}_1:x^2+y^2+2x+4y-20=0$ whose centre will be $\left(-1,-2\right)$ radius $r_1=\sqrt{1^2+2^2+20}=5$ and ${\mathrm{C}}_2:x^2+y^2+6x-8y+9=0$ whose centre is $\left(-3,4\right)$ and radius $r_2=\sqrt{9+16-9}=4$,

Now we cans see $C_1{C}_2=\sqrt{40}, r_1+r_2=9 \& r_1-r_2=1$

So, $r_1-r_2<C_1{C}_2<r_1+r_2$ so there will be $2$ common tangent,

Hence $n=2$ 

And length of tangent will be

${\left(r_1-r_2\right)}^2+l^2={\left(C_1{C}_2\right)}^2$

$\Rightarrow l=\sqrt{39}$

So, $\frac{l}{n^2}=\frac{\sqrt{39}}{4}$