1. We are given: $f(2)=4, \qquad f'(2)=1$

   We need to evaluate $L=\lim_{x\to 2}\frac{x^2f(2)-4f(x)}{x-2}.$

2. Substitute $f(2)=4$:

   =4\lim_{x\to 2}\frac{x^2-f(x)}{x-2}.$$

3. Since $f(2)=4=2^2$, the numerator becomes $0$ at $x=2$, so we can rewrite: $x^2-f(x)=(x^2-4)-(f(x)-4).$ Therefore, $\frac{x^2-f(x)}{x-2}=\frac{x^2-4}{x-2}-\frac{f(x)-4}{x-2}.$

4. Now take limits separately: $\lim_{x\to 2}\frac{x^2-4}{x-2} = \lim_{x\to 2}(x+2)=4,$ and $\lim_{x\to 2}\frac{f(x)-4}{x-2}=\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}=f'(2)=1.$

5. Hence, $\lim_{x\to 2}\frac{x^2-f(x)}{x-2}=4-1=3.$

6. Multiply by $4$: $L=4\cdot 3=12.$

7. Therefore, the required value is $\boxed{12}.$

8. Checking options:

   • A: $4$ ❌
   • B: $8$ ❌
   • C: $16$ ❌
   • D: $12$ ✅

So the correct option is D.