as $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x^3}=1$ non-zero finite

So, $d=e=f=0$

and $f\left(x\right)=x^3\left(x^3+a{x}^2+bx+c\right)$

Hence, $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x^3}=c=1$

Now, as $f\left(x\right)=x^6+a{x}^5+b{x}^4+x^3$

and $f^{\text{'}}\left(x\right)=0$ at $x=1$ and $x=-1$

i.e., $f^{\text{'}}\left(x\right)=6x^5+5a{x}^4+4b{x}^3+3x^2$

$f\text{'}\left(1\right)=0$

$\Rightarrow 6+5a+4b+3=0$

$\Rightarrow 5a+4b=-9$

and $f^{\text{'}}\left(-1\right)=0$

$\Rightarrow -6+5a-4b+3=0$

$\Rightarrow 5a-4b=3$

Solving both we get,

$a=\frac{-6}{10}=\frac{-3}{5}; b=\frac{-3}{2}$

$\therefore f\left(x\right)=x^6-\frac{3}{5}{x}^5-\frac{3}{2}{x}^4+x^3$

$\therefore 5f\left(2\right)=5\left[64-\frac{3}{5}·32-\frac{3}{2}·16+8\right]$

$=320-96-120+40$

$=144$