1. Since $p(x)$ is a polynomial of degree $4$ and has extrema at $x=1,2$, its derivative must vanish at these points.

   So, $p'(1)=0,\qquad p'(2)=0.$

2. A quartic polynomial has a cubic derivative. Since extrema are given at $x=1,2$, we can write $p'(x)=k(x-1)(x-2)(x-a)$ for some constants $k\neq 0$ and $a$.

3. Now use the limit condition: $\lim_{x\to 0}\left(1+\frac{p(x)}{x^2}\right)=2.$ This gives $\lim_{x\to 0}\frac{p(x)}{x^2}=1.$

   Since $p(x)$ is a polynomial, for this limit to be finite, we must have $p(0)=0 \quad \text{and} \quad p'(0)=0.$ Also the limit then equals $\frac{p''(0)}{2}=1 \implies p''(0)=2.$

4. Thus $x=0$ is a double root of $p(x)$. So we can write $p(x)=x^2(ax^2+bx+c).$

   Then $\frac{p(x)}{x^2}=ax^2+bx+c.$ Taking $x\to 0$, $c=1.$

   Hence $p(x)=x^2(ax^2+bx+1).$

5. Differentiate: $p(x)=ax^4+bx^3+x^2$ so $p'(x)=4ax^3+3bx^2+2x=x(4ax^2+3bx+2).$

6. Since extrema occur at $x=1$ and $x=2$, we need $p'(1)=0,\qquad p'(2)=0.$

   From $p'(1)=0$: 4a+3b+2=0.\tag{1}

   From $p'(2)=0$: 32a+12b+2=0 \implies 16a+6b+1=0.\tag{2}

7. Solve (1) and (2):

   Multiply (1) by $2$: 8a+6b+4=0.\tag{3}

   Subtract (3) from (2): $8a-3=0 \implies a=\frac{3}{8}.$

   Substitute into (1): $4\cdot \frac{3}{8}+3b+2=0$ $\frac{3}{2}+3b+2=0$ $3b=-\frac{7}{2}$ $b=-\frac{7}{6}.$

8. Now compute $p(2)$: $p(2)=2^2\left(a\cdot 2^2+b\cdot 2+1\right)$ $=4\left(4a+2b+1\right).$

   Substitute $a=\frac{3}{8}$ and $b=-\frac{7}{6}$: $4a=\frac{3}{2},\qquad 2b=-\frac{7}{3}.$

   So,

   =\frac{9-14+6}{6}=\frac{1}{6}.$$ Therefore, $$p(2)=4\cdot \frac{1}{6}=\frac{2}{3}.$$

9. But this seems inconsistent with the condition that $x=2$ is an extremum of the quartic. Let us instead use the structure of derivative more carefully.

   Since $p(x)=x^2(ax^2+bx+1)$, $p'(x)=x(4ax^2+3bx+2).$

   For both $x=1$ and $x=2$ to be extrema, the quadratic factor must be proportional to $(x-1)(x-2)$: $4ax^2+3bx+2 = \lambda(x-1)(x-2)=\lambda(x^2-3x+2).$

   Comparing constant terms: $2=2\lambda \implies \lambda=1.$

   Hence

   \qquad 3b=-3 \implies b=-1.$$

10. Therefore, $p(x)=x^2\left(\frac14 x^2-x+1\right).$

Now, $p(2)=4\left(\frac14\cdot 4-2+1\right)=4(1-2+1)=0.$

Thus, $\boxed{p(2)=0}.$