Given,

$f\left(x\right)=\frac{x^3}{3}+2bx+\frac{a{x}^2}{2}$ and $g\left(x\right)=\frac{x^3}{3}+ax+b{x}^2$

Now differentiating the above function we get,

$f^{\text{'}}\left(x\right)=x^2+2b+ax$

$g^{\text{'}}\left(x\right)=x^2+a+2bx$

Now given, $f^{\text{'}}\left(x\right)=0 and g^{\text{'}}\left(x\right)=0$ have a common extreme point,

So, $x^2+2b+ax=x^2+a+2bx$

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

$\therefore x=1$ is the common extreme point

Put $x=1$ in $f^{\text{'}}\left(x\right)=0$ or $g^{\text{'}}\left(x\right)=0$ we get,

$1+2b+a=0$

$\Rightarrow a+2b+7=6$