Given potential energy is

$U\left(x\right)=U_0\left(1-\text{cos}ax\right)$

Therefore force on the particle is given by

$F=-\frac{dU}{dx}=-U_{0}a\sin \left(\mathit{ax}\right)$

Using small angle approximation, $\sin (\mathit{ax})\approx ax$

Therefore $F=-U_0{a}^{2}x$

Time period of SHM is,

$T=2\pi \sqrt{\frac{m}{k}}$ where $k$ is force constant.

Therefore time period for given force is,

$T=2\pi \sqrt{\frac{m}{U_0{a}^2}}$