Two identical pith balls are charged by rubbing against one...

When the two identical pith balls are rubbed against each other, they acquire equal and opposite charges, leading to an electrostatic force of attraction between them.

The electrostatic force of attraction is given by Coulomb's law:
$F_e = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{r^2}$

Substituting the given values:
$F_e = \dfrac{9 \times 10^9 \times (2.0 \times 10^{-8})^2}{(0.03)^2} = \dfrac{9 \times 10^9 \times 4.0 \times 10^{-16}}{9 \times 10^{-4}} = 4.0 \times 10^{-3} \text{ N}$

Let $\theta$ be the angle made by each string with the vertical. The suspension points are $5 \text{ cm}$ apart, and the balls are $3 \text{ cm}$ apart in equilibrium. By symmetry, the horizontal displacement of each ball from the vertical line passing through its suspension point is:
$x = \dfrac{5 \text{ cm} - 3 \text{ cm}}{2} = 1 \text{ cm}$

The length of each string is $L = 20 \text{ cm}$. Thus, we have:
$\sin\theta = \dfrac{x}{L} = \dfrac{1}{20} = 0.05$

The cosine of the angle is:
$\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - (0.05)^2} = \sqrt{0.9975} \approx 0.9987$

In equilibrium, the horizontal component of the tension balances the electrostatic force, and the vertical component balances the weight of the ball:
$T \sin\theta = F_e$
$T \cos\theta = mg$

From the first equilibrium condition, the tension $T$ in the string is:
$T = \dfrac{F_e}{\sin\theta} = \dfrac{4.0 \times 10^{-3}}{0.05} = 8.0 \times 10^{-2} \text{ N}$

From the second equilibrium condition, the mass $m$ of each ball is:
$m = \dfrac{T \cos\theta}{g}$

Taking $g = 9.8 \text{ m/s}^2$:
$m = \dfrac{8.0 \times 10^{-2} \times 0.9987}{9.8} \approx 8.15 \times 10^{-3} \text{ kg} = 8.15 \text{ g}$

Rounding to two significant figures, the mass is approximately $8.2 \text{ g}$.

Answer: $8.2 \text{ g}$, $8.0 \times 10^{-2} \text{ N}$