Given that, $ f(x+y) = f(x) + f(y) , f(1) = 2 $
And $, f(a+1) + f(a+2) + f(a+3) + \dots + f(a+n) = 16(2^{n}-1) \quad \longrightarrow (1) $
Now, put $n=1$, we get
$ f(a+1) = 16(2-1) = 16 $
$ \Rightarrow f(a) f(1) = 16 $
$ \Rightarrow f(a) \cdot 2 = 16 $
$ \Rightarrow f(a) = 8 \quad \longrightarrow (2) $
And, put $n=2$, we get
$ f(a+1) + f(a+2) = 16(2^{2} – 1) $
$ \Rightarrow f(a) f(1) + f(a) f(2) = 48 $
$ \Rightarrow 8 \cdot 2 + 8 \cdot f(2) = 48 $
$ \Rightarrow 16 + 8 f(2) = 48 $
$ \Rightarrow 8 f(2) = 32 $
$ \Rightarrow f(2) = 4 \quad \longrightarrow (3) $
And, put $n=3$, we get
$ f(a+1) + f(a+2) + f(a+3) = 16(2^{3}-1) $
$ \Rightarrow f(a) f(1) + f(a) f(2) + f(a) f(3) = 16(7) $
$ \Rightarrow 8 \cdot2 + 8 \cdot 4 + 8 f(3) = 112 $
$ \Rightarrow 16 + 32 + 8 f(3) = 112 $
$ \Rightarrow 48 + 8 f(3) = 112 $
$ \Rightarrow 8 f(3) = 64 $
$ \Rightarrow f(3) = 8 \quad \longrightarrow (4) $
On comparing equation $(2)$ and $(4)$, we get
$ \boxed{a=3} $
Correct Answer $: 3 $