Answer
a) $75.4\;\rm nA$
b) $133\;\rm s$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the current is given by the amount of charge moving at a definite interval of time,
$$I=\dfrac{Q}{t}\tag 1$$
We know that the film rolls at a speed of 90 rpm (revolution per minute), so its linear speed is
$$v=\omega r$$
Plug the known, but we need to convert all the units to SI system units.
$$v=\rm\left(\dfrac{90\; rev}{60\;s}\right) \left(\dfrac{2\pi\; rad}{1\;rev}\right) \left(1\;cm\right)=\bf 0.09425\;\rm m/s\tag 2$$
Recalling that $v=x/t$, so
$$x=vt$$
We are told that the plastic piece has a uniform surface charge density, so
$$\eta=\dfrac{Q}{A}$$
Hence,
$$Q=A\eta $$
where $A=0.04 x$ where $x$ is the distance moved by the film that unwraps it by the same distance, and 0.04 m is the width of the film.
Hence,
$$Q=0.04 x\eta \tag 3$$
Plug into (1),
$$I=\dfrac{0.04 x\eta}{t} $$
where $x/t=v$, so
$$I= 0.04 v\eta $$
Plug the known;
$$I= (0.04\;\rm m) \left(\dfrac{0.09425\;\rm m}{1\;\rm s}\right)\left(\dfrac{|-2|\times 10^{-9}\;\rm C}{1\;\rm cm^2}\right)\left(\dfrac{100\;\rm cm}{1\;\rm m}\right)^2 $$
$$I=\color{red}{\bf 75.4}\;\rm nA$$
$$\color{blue}{\bf [b]}$$
Solving (1) for $t$ to find how long it takes the roller to accumulate a charge of -10 $\mu$C.
$$t=\dfrac{Q_1}{I}$$
Plug the known;
$$t=\dfrac{|-10\times 10^{-6}|}{(75.4\times 10^{-9})}$$
$$t=\color{red}{\bf 133}\;\rm s$$
