Let $f\left(x\right)$   be a polynomial of degree 6, which satisfies  $\underset{x\to 0}{\lim }{(1+\frac{f\left(x\right)}{x^3})}^{1/x}=e^2$ and has local maximum at $x=1$  and local minimum at $x=0$  and 2 then $5f\left(3\right)$  is equal to……

The given limit will exist only if there is no constant term, one degree, two degree and three degree terms in $f\left(x\right).$   
$\Rightarrow f\left(x\right)=a{x}^4+b{x}^5+c{x}^6.$  In this case, the given limit can be written as
$\underset{x\to 0}{\lim }{(1+ax+b{x}^2+c{x}^3)}^{1/x}\Rightarrow e^{\underset{x\to 0}{Lt}(a+bx+c{x}^2)}=e^a=e^2\Rightarrow a=2$ 
Thus,  $f\left(x\right)=2x^4+b{x}^5+c{x}^6$
$\Rightarrow f\text{'}\left(x\right)=x^3(8+5bx+6c{x}^2)$
Since, $f\left(x\right)$  has local extremum at  $x=0,1,2.$
So   $f\text{'}\left(0\right)=f\text{'}\left(1\right)=f\text{'}\left(2\right)=0.\Rightarrow 8+5b+6c=0,8+10b+24c=0$ 
$\Rightarrow c=2/3,b=-12/5.$ Hence,  $f\left(x\right)=2x^4-\frac{12}{5}{x}^5+\frac{2}{3}{x}^6.\Rightarrow f\left(3\right)=\frac{324}{5}$