Two identically charged particles are attached to the opposite ends...

Let the natural length of the spring be $L$ and the extension be $x$.

The electrostatic repulsive force between the two charges is given by Coulomb's law:

$F_{e} = \dfrac{1}{4\pi \epsilon_{0}} \dfrac{q^{2}}{(L+x)^{2}}$

Since the extension $x$ is small compared to the natural length $L$, $L+x \approx L$. Thus,

$F_{e} \approx \dfrac{1}{4\pi \epsilon_{0}} \dfrac{q^{2}}{L^{2}}$

The restoring force of the spring is:

$F_{s} = kx$

In equilibrium, the electrostatic repulsive force is balanced by the restoring force of the spring:

$kx = \dfrac{1}{4\pi \epsilon_{0}} \dfrac{q^{2}}{L^{2}}$

Substituting the given values, $q = 2.0 \times 10^{-8} \text{ C}$, $L = 10 \text{ cm} = 0.1 \text{ m}$, $k = 100 \text{ N m}^{-1}$, and $\dfrac{1}{4\pi \epsilon_{0}} = 9 \times 10^{9} \text{ N m}^{2} \text{ C}^{-2}$:

$100 x = \dfrac{9 \times 10^{9} \times (2.0 \times 10^{-8})^{2}}{(0.1)^{2}}$

$100 x = \dfrac{9 \times 10^{9} \times 4.0 \times 10^{-16}}{0.01}$

$100 x = \dfrac{36 \times 10^{-7}}{0.01}$

$100 x = 36 \times 10^{-5}$

$x = 3.6 \times 10^{-6} \text{ m}$