For SHM with x(t)=Asin(ωt), the potential energy is
U=21mω2x2=21mω2A2sin2(ωt).
Maximum potential energy occurs when sin2(ωt)=1, i.e., when ωt=2π,23π,...
The first maximum occurs at ωt=2π, giving t=2ωπ=2π⋅2πT=4T.
Since this is given as t=2βT.
We have 4T=2βT, so β=2.