Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the current is given by
$$I=\dfrac{dQ}{dt}$$
where $I=2 e^{-t/2\mu}$, so
$$2 e^{-t/2\mu}=\dfrac{dQ}{dt}$$
Hence,
$$dQ=2 e^{-t/2\mu }dt$$
Taking the integral for both sides;
$$\int_0^QdQ=2 \int_0^t e^{-t/2\mu} dt$$
$$ Q=2 (-2\mu) e^{-t/2\mu} \bigg|_0^t $$
$$ Q=-4\mu ( e^{-t/2\mu}-1) $$
$$\boxed{ Q= 4\times 10^{-6} \left(1-e^{\left[\frac{-t}{2\times 10^{-6} }\right]} \right)} $$
$$\color{blue}{\bf [b]}$$
See the graph below,