Given $f\left(x+y\right)=2f\left(x\right)·f\left(y\right)$ & $f\left(1\right)=2$

Now putting $x=1$ & $y=1$  in $f\left(x+y\right)=2f\left(x\right)·f\left(y\right)$ we get

$f\left(1+1\right)=2f\left(1\right)·f\left(1\right)=2\times 2^2=2^3$

So $f\left(2\right)=2^3$, Similarly

$f\left(3\right)=2^5, f\left(4\right)=2^7.....$

Now

$\sum _{k=1}^{10}f\left(\alpha +k\right)=\sum _{k=1}^{10}2f\left(\alpha \right)·f\left(k\right)=\frac{512}{3}\left(2^{20}-1\right)$

$\Rightarrow 2f\left(\alpha \right)\sum _{k=1}^{10}f\left(k\right)=\frac{512}{3}\left(2^{20}-1\right)$ 

$\Rightarrow 2f\left(\alpha \right)\left\{f\left(1\right)+f\left(2\right)\cdots f\left(10\right)\right\}=\frac{512}{3}\left(2^{20}-1\right)$

$\Rightarrow 2f\left(\alpha \right)\left(2+2^3+2^5\dots \right)=\frac{512}{3}\left(2^{20}-1\right)$

$\Rightarrow 2f\left(\alpha \right)\frac{2\left({\left(2^2\right)}^{10}-1\right)}{2^2-1}=\frac{512}{3}\left(2^{20}-1\right)$

$\Rightarrow 2\times 2f\left(\alpha \right)\times \frac{2^{20}-1}{3}=\frac{512}{3}\left(2^{20}-1\right)$

Now comparing both side we get

$4f\left(\alpha \right)=512\Rightarrow f\left(\alpha \right)=128$

$\Rightarrow f\left(\alpha \right)=2^7\Rightarrow \alpha =4$.