Question

the equation I = 2.4e-6t defines how current varies with time

Answer 1

$$I=\dfrac{dq}{dt}=2.4e^{-6t}$$

The quantity of charge which has to be passed through a cross section of the wire during the time $t_1$ to $t_2$ is​

$$\displaystyle\int_{t_1}^{_2}2.4e^{-6t}dt=[-0.4e^{-6t}]\begin{matrix} t_2 \\ t_1 \end{matrix}$$

$$=-0.4e^{-6t_2}+0.4e^{-6t_1}$$