Answer
$T=2\pi\sqrt{\frac{mK_{\circ}^2}{C}}$
Work Step by Step
We can determine the natural period of oscillation as follows:
$\omega_n=\sqrt{\frac{K_{eq}}{m_{eq}}}$
$\implies \omega_n=\sqrt{\frac{C}{mK_{\circ}^2}}$
Now,
$T=\frac{2\pi}{\omega_n}$
Thus:
$\implies T=2\pi\sqrt{\frac{mK_{\circ}^2}{C}}$
