Answer
$F=-2.56 \times 10^{-10} \mathrm{~N} .$
Work Step by Step
The electrostatic force on the grain by the bee is
$$
F=\frac{k Q q}{(d+D / 2)^2}+\frac{k Q(-q)}{(D / 2)^2}\\=-k Q|q|\left[\frac{1}{(D / 2)^2}-\frac{1}{(d+D / 2)^2}\right]
$$
where $D=1.000 \mathrm{~cm}$ is the diameter of the sphere representing the honeybee, and $d=40.0 \mu \mathrm{m}$ is the diameter of the grain. Substituting the values, we obtain
$$
\begin{aligned}
F & =-\left(8.99 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{C}^2\right)\left(45.0 \times 10^{-12} \mathrm{C}\right)\left(1.000 \times 10^{-12} \mathrm{C}\right)\left[\frac{1}{\left(5.00 \times 10^{-3} \mathrm{~m}\right)^2}-\frac{1}{\left(5.04 \times 10^{-3} \mathrm{~m}\right)^2}\right] \\
& =-2.56 \times 10^{-10} \mathrm{~N} .
\end{aligned}
$$
The negative sign implies that the force between the bee and the grain is attractive. The magnitude of the force is $|F|=2.56 \times 10^{-10} \mathrm{~N}$.
