Given,

$\frac{6}{3^{12}}+\frac{10}{3^{11}}+\frac{20}{3^{10}}+\frac{40}{3^9}+\dots ..+\frac{10240}{3}=2^n·m$

On rearranging we get, 

$\Rightarrow \frac{1}{3^{12}}+5\left(\frac{2^0}{3^{12}}+\frac{2^1}{3^{11}}+\frac{2^2}{3^{10}}+\dots \dots +\frac{2^{11}}{3}\right)=2^n·m$

$\Rightarrow \frac{1}{3^{12}}+5\left(\frac{1}{3^{12}}\frac{\left({\left(6\right)}^2-1\right)}{\left(6-1\right)}\right)=2^n·m$

$\Rightarrow \frac{1}{3^{12}}+\frac{5}{5}\left(\frac{1}{3^{12}}·2^{12}·3^{12}-\frac{1}{3^{12}}\right)=2^n·m$

$\Rightarrow \frac{1}{3^{12}}+2^{12}-\frac{1}{3^{12}}=2^n·m$

$\Rightarrow 2^n·m=2^{12}$

On comparing both side we get,

$\Rightarrow m=1$ and $n=12$

So, $m·n=12$