Given functional equation is 

$f\left(m+n\right)=f\left(m\right)+f\left(n\right);m,n\in N ....\left(1\right)$

Put $m=n=3$

$f(3+3)=f\left(3\right)+f\left(3\right)$

$\Rightarrow f\left(3\right)=9\left[\because f\left(6\right)=18\right]$

Put $m=2,n=1 \text{in equation} \left(1\right)$

$\therefore f\left(3\right)=f(2+1)=f\left(2\right)+f\left(1\right)$

$=f(1+1)+f\left(1\right)$

$=f\left(1\right)+f\left(1\right)+f\left(1\right)$

$9=3f\left(1\right)$

$\Rightarrow f\left(1\right)=3$

$\therefore f\left(2\right)=f(1+1)=f\left(1\right)+f\left(1\right)=6$

Now,

$f\left(2\right)·f\left(3\right)=\left(6\right)\left(9\right)=54$