Answer
$4.16\times 10^{-4}\;\rm N$
Work Step by Step
Let's assume that the charge is uniformly distributed along the plastic stirrer. So its linear charge density is then $\lambda_s$, which is different from the linear charge density of the infinite charged wire $\lambda$.
We know that the electric field of an infinite line of charge at a distance $x$ from its axis is given by
$$E=\dfrac{1}{4\pi \epsilon_0}\dfrac{2\lambda}{x}\;\hat i\tag 1$$
So the small force exerted by this electric field on the small fragment of the stirrer which is at distance $x$, as we see in the figure below, is given by
$$dF=Edq\tag 2$$
where $dq$ is given by
$$\lambda_1=\dfrac{dq}{dx}=\dfrac{Q_1}{L}$$
Hence,
$$dq=\dfrac{Q_1 dx}{L}$$
Plug into (2), and plug $E$ from (1),
$$dF=\dfrac{1}{4\pi \epsilon_0}\dfrac{2\lambda}{x}\dfrac{Q_1 dx}{L} $$
$$dF=\dfrac{2Q_1\lambda}{(4\pi \epsilon_0)L} \dfrac{ dx}{x} $$
To find the full force exerted on the whole plastic stirrer, we need to integrate from $x=2$ cm to $x=2+6=8$ cm.
$$\int_0^F dF=\dfrac{2Q_1\lambda}{(4\pi \epsilon_0)L} \int_{x=0.02}^{x=0.08}\dfrac{ dx}{x} $$
$$ F=\dfrac{2Q_1\lambda}{(4\pi \epsilon_0)L} \ln x\bigg| _{x=0.02}^{x=0.08} $$
Plug the known;
$$ F=\dfrac{2(10\times 10^{-9})(1\times 10^{-7}) (9\times 10^9)}{ 0.06} (\ln 0.08-\ln0.02) $$
$$F=\color{red}{\bf 4.16\times 10^{-4}}\;\rm N$$
