Given that, $f(ab)=f(a)f(b);\forall$ $a,b \in \mathbb{Z}^{+}.$

Now, $f(1\cdot1)=f(1)f(1)$

$\Rightarrow f(1)=f^{2}(1)$

$\Rightarrow f^{2}(1)-f(1)=0$

$\Rightarrow f(1)(f(1)-1)=0$

$\Rightarrow f(1)=0; f(1)=1$

$\therefore$ The largest possible value of $f(1)$ is $1.$

Correct Answer $:\text{A}$