1. The charge flowing through a wire in a time interval is given by

$Q=\int_{t_1}^{t_2} I(t)\,dt$

Here,

$I(t)=0.02t+0.01\ \text{A}$

and the interval is from $t=1\,\text{s}$ to $t=2\,\text{s}$.

2. Substitute into the integral:

$Q=\int_1^2 (0.02t+0.01)\,dt$

3. Integrate term by term:

$\int (0.02t+0.01)dt = 0.02\cdot \frac{t^2}{2}+0.01t = 0.01t^2+0.01t$

4. Apply the limits from $1$ to $2$:

$Q=\left[0.01t^2+0.01t\right]_1^2$

At $t=2$:

$0.01(2)^2+0.01(2)=0.04+0.02=0.06$

At $t=1$:

$0.01(1)^2+0.01(1)=0.01+0.01=0.02$

So,

$Q=0.06-0.02=0.04\ \text{C}$

5. Therefore, the correct option is:

$\boxed{\text{D: }0.04\ \text{C}}$

6. Comparison with stored answer:

Stored correct answer is D, which matches the derived answer.