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

Put $x=y=1$

$\Rightarrow f\left(2\right)=f\left(1\right)·f\left(1\right)=2^2$

Now, put $x=1, y=2$

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

Now, put $x=1, y=3$

$\Rightarrow f\left(4\right)=f\left(1\right)·f\left(3\right)=2^4$

Similarly, $f\left(10\right)=2^{10}$

Now $\sum _{k=1}^{10} f\left(a+k\right)=16\left(2^{10}-1\right)$

$\Rightarrow \sum _{k=1}^{10}f\left(a\right)f\left(k\right)=16\left(2^{10}-1\right)$

$\Rightarrow f\left(a\right)\sum _{k=1}^{10}f\left(k\right)=16\left(2^{10}-1\right)$

$\Rightarrow f\left(a\right)\left(f\left(1\right)+f\left(2\right)\dots +f\left(10\right)\right)=16\left(2^{10}-1\right)$

$\Rightarrow f\left(a\right)\left(2^1+2^2+2^3+\dots +2^{10}\right)=16\left(2^{10}-1\right)$

Using the sum of $n$ terms of a geometric progression, i.e. $a+ar+a{r}^2+a{r}^3+...+a{r}^{n-1}=\frac{a\left(r^n-1\right)}{r-1},$ we get

$f\left(a\right)\frac{2\left(2^{10}-1\right)}{2-1}=16\left(2^{10}-1\right)$

$\Rightarrow f\left(a\right)=8=2^3$

Also, we know that $f\left(3\right)=8$

$\Rightarrow f\left(a\right)=f\left(3\right)$

$\Rightarrow a=3.$