Let $f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e$

Then $\underset{x\to 0}{\lim }\left[1+\frac{f\left(x\right)}{x^2}\right]=3$

$\Rightarrow \underset{x\to 0}{\lim }\left[1+a{x}^2+bx+c+\frac{d}{x}+\frac{e}{x^2}\right]=3$

This limit exists when $d=e=0$.

So,  $\underset{x\to 0}{\lim }\left[1+a{x}^2+bx+c\right]=3$

$\Rightarrow c+1=3$

$\Rightarrow c=2$

It is given, $x=1$ and $x=2$ are the solution of $f^{\text{'}}\left(x\right)=0$.

$\Rightarrow f^{\text{'}}\left(x\right)=4a{x}^3+3b{x}^2+2cx$

$\Rightarrow x\left(4a{x}^2+3bx+2c\right)=0$

Given $1, 2$ are the roots of the quadratic equation. 

$\Rightarrow \text{Sum of roots}=\frac{-3b}{4a}=1+2=3$

$\Rightarrow b=-4a$

Product of roots $=\frac{2c}{4a}=1·2=2$

$\Rightarrow a=\frac{c}{4}$

$\Rightarrow a=\frac{1}{2}\text{, }b=-2$

$\therefore f\left(x\right)=\frac{1}{2}{x}^4-2x^3+2x^2$

$\Rightarrow f\left(2\right)=8-16+8$

$\Rightarrow f\left(2\right)=0$