$\text{ Let }{f}^{\mathrm{\prime }}\left(x\right)\text{ be continuous at }x=0\text{ and }{f}^{\prime \prime }\left(0\right)=4\text{. The value of }$$\underset{x\to 0}{\lim }\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$

$\begin{array}{l}\underset{x\to 0}{\lim }\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}\left[Form\frac{0}{0}\right]\\ =\underset{x\to 0}{\lim }\frac{2f\text{'}\left(x\right)-6f\text{'}\left(2x\right)+4f\text{'}\left(4x\right)}{2x}\left[Form\frac{0}{0}\right]\\ =\underset{x\to 0}{\lim }2f\text{'}\text{'}\left(x\right)-12f"\left(2x\right)+16f"\left(4x\right)\\ =f"\left(0\right)-6f"\left(0\right)+gf"\left(0\right)=3f"\left(0\right)=3\times 4=12\end{array}$

$\text{Hence\hspace{0.17em}(3)is\hspace{0.17em}the\hspace{0.17em}correct\hspace{0.17em}answer.}$