Question

Suppose you have two small pith balls that are 5.5 cm apart and have equal charges of -29 nc?

Answer 1

The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.

The Coulomb force between the two balls is:
$F= k_e \frac{ q_1 q_2}{r^2}$
where $k_e=8.99\cdot10^9~N m^2 C^{-2}$ is the Coulomb's constant, $q_1=q_2=29~nC=29\cdot 10^{-9}~C$ is the intensity of the two charges, and $r=5.5~cm=0.055~m$ is the distance between them.

Substituting these numbers into the equation, we get
$F=2.5~10^{-3}~N$

The force is repulsive, because the charges have same sign and so they repel each other.