1. Let the polynomial be

Since $f(x)$ is a monic polynomial of degree $6$, write $f(x)=x^6+ax^5+bx^4+cx^3+dx^2+ex+g.$

2. Use the limit condition

Given $\lim_{x\to 0}\frac{f(x)}{x^3}=1.$ For this limit to exist and be finite, all terms of degree less than $3$ must vanish: $g=0,\quad e=0,\quad d=0.$ Also the coefficient of $x^3$ must be $1$, so $c=1.$ Hence, $f(x)=x^6+ax^5+bx^4+x^3.$

3. Use the extrema at $x=-1$ and $x=1$

If $x=\pm1$ are extrema, then $f'(1)=0,\qquad f'(-1)=0.$ Now, $f'(x)=6x^5+5ax^4+4bx^3+3x^2.$

So at $x=1$, $6+5a+4b+3=0$ $5a+4b+9=0 \quad ...(1)$

At $x=-1$, $-6+5a-4b+3=0$ $5a-4b-3=0 \quad ...(2)$

4. Solve for $a,b$

Adding (1) and (2): $10a+6=0$ $a=-\frac{3}{5}.$

Substitute into (2): $5\left(-\frac35\right)-4b-3=0$ $-3-4b-3=0$ $-4b=6$ $b=-\frac32.$

Thus, $f(x)=x^6-\frac35x^5-\frac32x^4+x^3.$

5. Find $f(2)$

$f(2)=2^6-\frac35(2^5)-\frac32(2^4)+2^3$ $=64-\frac35\cdot 32-\frac32\cdot 16+8$ $=64-\frac{96}{5}-24+8$ $=48-\frac{96}{5}$ $=\frac{240-96}{5}$ $=\frac{144}{5}.$

Therefore, $5f(2)=144.$

6. Comparison with stored answer

Derived answer is $144$, which matches the stored correct answer.