Let

$\because f(x+y)=f(x) \cdot f(y)$

$\Rightarrow$ Let $f(x)=t^{x}$

$\because \mathrm{f}(1)=2$

$\therefore t=2$

$\Rightarrow \mathrm{f}(\mathrm{x})=2^{\mathrm{x}}$

Since, $\sum_{k=1}^{10} f(a+k)=16\left(2^{10}-1\right)$

Then, $\sum_{k=1}^{10} 2^{a+k}=16\left(2^{10}-1\right)$

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

$\Rightarrow 2^{a} \times \frac{\left(\left(2^{10}\right)-1\right) \times 2}{(2-1)}=16 \times\left(2^{10}-1\right) \Rightarrow 2.2^{a}=16$

$\Rightarrow a=3$