Answer
$+40\;\rm nC$
Work Step by Step
We know that the electric field exerted on a point at distance $r$ from the center of a long charged rod is given by
$$E=\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q|}{r\sqrt{r^2+\left(\frac{L}{2}\right)^2}}$$
Thus the repulsive force between the bead and the rod at $r$ distance is given by
$$F=|q|E=\dfrac{1}{4\pi \epsilon_0}\dfrac{|q||Q|}{r\sqrt{r^2+\left(\frac{L}{2}\right)^2}}$$
Hence, the total charge in the rod is given by
$$|Q|=\dfrac{4\pi \epsilon_0 rF\;\sqrt{r^2+\left(\frac{L}{2}\right)^2} }{|q|}$$
Plugging the known;
$$|Q|=\dfrac{4\pi (8.85\times 10^{-12}) (0.04)(840\times 10^{-6})\;\sqrt{0.04^2+\left(\frac{0.10}{2}\right)^2} }{|6\times 10^{-9}|}=\bf\pm 40\;\rm nC$$
And since the force between the bead and the rod is a repulsive force, then the charge of the rod must be positive.
$$Q=\color{red}{\bf +40}\;\rm nC$$