We have, $f\left(x+y\right)=f\left(x\right)+2y^2+kxy$ for all $x,y\in R$

$\Rightarrow \frac{f\left(x+y\right)-f\left(x\right)}{y}=2y+kx$ for all $x\in R$

$\Rightarrow \underset{y\to 0}{\lim }\frac{f\left(x+y\right)-f\left(x\right)}{y}=\underset{y\to 0}{\lim }\left(2y+kx\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=kx$ for all $x\in R$

$\Rightarrow f\left(x\right)=\frac{k{x}^2}{2}+C$ for all $x\in R$  [by integration]

But $f\left(1\right)=2$ and $f\left(2\right)=8$

$\therefore 2=\frac{k}{2}+C$ and $8=2k+C$

$k=4$ and $C=0$

Hence, $f\left(x\right)=2x^2$ for all $x\in R$

So, option (a) is correct.