Let $a>2,a\in N$  be a constant. If there are just $18$  positive integers satisfying the inequality $\left(x-a\right)\left(x-2a\right)\left(x-a^2\right)<0,$  then the value of$a$  is

As $a>2,$  hence $a^2>2a>a>2$

Now $\left(x-a\right)\left(x-2a\right)\left(x-a^2\right)<0$ $\Rightarrow$  the solution set is as

Between $\left(0,a\right)$  there are $\left(a,-1\right)$  positive integers

Between $\left(2a,a^2\right)$  there are $\left(a^2-2a-1\right)$  positive integers

$\therefore a^2-2a-1+a-1=18$ $\Rightarrow a^2-a-20=0$ $\Rightarrow \left(a-5\right)\left(a+4\right)=0$

$\therefore a=5anda=-4$  but $a>2$ $\therefore a=5$