For any quadratic polynomial $f\left(x\right),$ it is true that
$f\left(x\right)=f\left(a\right)+f\text{'}\left(a\right)(x-a)+\frac{f"\left(a\right)}{2!}{\left(x-a\right)}^2$ where $a$ is any real number.
If $\frac{3x^2+4x+7}{(x-2{)}^3}=\frac{A}{(x-2{)}^3}+\frac{B}{(x-2{)}^2}+\frac{C}{(x-2)}$ and $g\left(x\right)=3x^2+4x+7$ then $A+B+C=$

$\frac{3x^2+4x+7}{(x-2{)}^3}=\frac{A}{(x-2{)}^3}+\frac{B}{(x-2{)}^2}+\frac{C}{(x-2)}$ is the given equation, which can be written as,

$3x^2+4x+7=A+B(x-2)+C(x-2{)}^2$

$3x^2+4x+7=A+B(x-2)+C(x^2-4x+4)$

Comparing the coefficients we get,

$$\begin{gathered} A-2B+4C=7 \text{(coefficients of constants)} \\ B-4C=4 \text{(coefficients of }x\text{)} \\ C=3 \text{(coefficients of }{x}^2\text{)} \end{gathered}$$

Solving the equations we get, $A=3,B=4,C=16$

$A+B+C=46$

from given,

$g\left(x\right)=3x^2+4x+7 \Rightarrow g\left(2\right)=27$

$g\text{'}\left(x\right)=6x+4\Rightarrow g\text{'}\left(2\right)=16$

$g\text{'}\text{'}\left(x\right)=6$

Comparing from options, $g\left(2\right)+g\text{'}\left(2\right)+\frac{g"\left(2\right)}{2!}=46$

Hence, it is the required answer.