1. Given current-time relation

The current is given as $i(t)=\alpha t+\beta t^2$ with $\alpha=20\ \text{A s}^{-1}, \qquad \beta=8\ \text{A s}^{-2}$

We need the total charge crossing the wire section in time $0$ to $15\,\text{s}$.

2. Use relation between charge and current

Charge is the time integral of current: $q=\int_0^{15} i(t)\,dt$

So, $q=\int_0^{15}(\alpha t+\beta t^2)\,dt$

Substitute the given values: $q=\int_0^{15}(20t+8t^2)\,dt$

3. Integrate term by term

$q=20\int_0^{15} t\,dt+8\int_0^{15} t^2\,dt$

Using $\int t\,dt=\frac{t^2}{2}, \qquad \int t^2\,dt=\frac{t^3}{3}$

we get $q=20\left[\frac{t^2}{2}\right]_0^{15}+8\left[\frac{t^3}{3}\right]_0^{15}$

4. Evaluate at limits

First term: $20\cdot \frac{15^2}{2}=20\cdot \frac{225}{2}=2250$

Second term: $8\cdot \frac{15^3}{3}=8\cdot \frac{3375}{3}=8\cdot 1125=9000$

Therefore, $q=2250+9000=11250\,\text{C}$

5. Match with options

$\boxed{q=11250\,\text{C}}$

So the correct option is: D: 11250 C