Given potential energy of oscillating mass at displacement x is
U(x)=10-10cos6x
So periodic force (F(x)) acting on the oscillating mass of 1kg will be
F(x)= -dU/dx= -10×6sin6x
For small oscillation sin6x =6x
So F(x)= -360x
So periodic acceleration
a(x)=F(x)/1=360x
Now by definition of SHM we know that periodic acceleration a(x) on the oscillating mass is related with w the angular frequency of the reference point associated with SHM
a(x)=-w^2x,
So w^2 = 360
=>w= 6√10
=>4π^2/T^2=360
=>T^2=π^2/90
=>T^2=10/90=1/9 (inserting π^2=10)
=>T=1/3 s