Oscillating electron

Consider that the electron is $x$ distance away from the sphere.

Now consider a spherical Gaussian surface of radius $x$ .

Applying Gauss's law, we have

$\phi=\frac {q_{enclosed}}{\epsilon_{0}}$

$\displaystyle\int E \cdot dA = \frac {q_{enclosed}}{\epsilon_{0}}$

$E \cdot 4 \pi x^2 = \frac{\frac {4}{3} \pi x^3\cdot \rho}{\epsilon_{0}}$

So, the electric field $E=\frac{\rho x}{3\epsilon_0}$

Hence, the force on the electron when it is $x$ distance away from the center of the sphere

= $-e \cdot E = \frac{-e \rho x}{3 \epsilon_0}$

Now, $F=ma$ .

So,

$a=\frac{F}{m}= \frac{-e \rho x}{3 \epsilon_0 m_{e}}$

Hence, the electron's motion is simple harmonic.

Comparing $a=-\omega^2 x$ and $a= \frac{-e \rho x}{3 \epsilon_0 m_{e}}$ ,

$\omega = \sqrt{\frac{e \rho}{3 \epsilon_0 m_{e}}}$

$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{3 \epsilon_0 m_{e}}{e \rho}}$

Substituting the values

$m_e=9.1 \times 10^{-31}kg$

$\epsilon_0=8.85 \times 10^{-12}\frac{F}{m}$

$e$ (charge on electron) $=1.6 \times 10^{-19}C$

And $\rho=2 \times 10^{-7}\frac{C}{m^3}$ ,

We get $T=\boxed{0.169 \mu s}$