Given $f\left(x\right)=2x^3-9a{x}^2+12a^{2}x+1$.

Differentiating $f\left(x\right)$ with respect to $x$, we get

$f^{\text{'}}\left(x\right)=6x^2-18ax+12a^2$

Again, differentiating $f^{\text{'}}\left(x\right)$ with respect to $x$, we get

$f^{"}\left(x\right)=12x-18a$

For maximum or minimum, $f^{\text{'}}\left(x\right)=0$.

$\Rightarrow 6x^2-18ax+12a^2=0$

$\Rightarrow x^2-3ax+2a^2=0$

$\Rightarrow \left(x-a\right)\left(x-2a\right)=0$

$\Rightarrow x=a$or $x=2a$

Now, $f^{"}\left(a\right)=12a-18a=-6a<0$ and $f^{"}\left(2a\right)=24a-18a=6a>0$.

Therefore, $f\left(x\right)$ is maximum at $x=a$ and minimum at $x=2a$.

Thus, $f\left(x\right)$ is maximum at $x=a$ and minimum at $x=2a$.

Hence, $p=a$ and $q=2a$.

Given $p^2=q$.

$\Rightarrow a^2=2a$

$\Rightarrow a\left(a-2\right)=0$

$\Rightarrow a=2 \text{or} a=0$

But $a>0$.

So, $a=2$.