1. Find the point of intersection in the second quadrant

The curve is $x^2=2y \implies y=\frac{x^2}{2}$

The line is $y-2x-6=0 \implies y=2x+6$

At intersection: $\frac{x^2}{2}=2x+6$

Multiply by $2$: $x^2=4x+12$ $x^2-4x-12=0$ $(x-6)(x+2)=0$

So, the intersection points have $x=6 \quad \text{or} \quad x=-2$

Since we need the point in the second quadrant, we take $x=-2$ Then $y=2(-2)+6=2$

Hence, $(a,b)=(-2,2)$

2. Write the integral

Given $I=\int_a^b \frac{9x^2}{1+5^x}\,dx$

Since $a=-2$ and $b=2$, $I=\int_{-2}^{2} \frac{9x^2}{1+5^x}\,dx$

3. Use the symmetry trick

Let $f(x)=\frac{9x^2}{1+5^x}$

Then $f(-x)=\frac{9x^2}{1+5^{-x}}$

Now, $\frac{1}{1+5^{-x}}=\frac{5^x}{1+5^x}$

So, $f(-x)=\frac{9x^2\,5^x}{1+5^x}$

Therefore, $f(x)+f(-x)=9x^2\left(\frac{1}{1+5^x}+\frac{5^x}{1+5^x}\right)=9x^2$

Using $\int_{-a}^{a} f(x)\,dx=\int_0^a [f(x)+f(-x)]\,dx,$ we get $I=\int_0^2 9x^2\,dx$

4. Evaluate the integral

$I=9\int_0^2 x^2\,dx=9\left[\frac{x^3}{3}\right]_0^2$ $=9\cdot \frac{8}{3}=24$

5. Check options

The value is $\boxed{24}$

So the correct option is C.

6. Comparison with stored answer

Stored correct answer: C

Derived answer: C

They agree.