The reaction is second-order in [A].
For the reaction A ⟶ products
The rate law is r = k[A]^m
Consider two Experiments 1 and 2.
r₂/r₁ = {k[A]₂^m}/{k[A]₁^m }
r₂/r₁ = {[A]₂/[A]₁}^m
______________
r₂/r₁ = {(⅓[A])/[A]}^m = (⅑r₁)/r₁
(⅓)^m= ⅑
3^m = 9
m = 2
The reaction is second-order in [A].
If dividing the concentration by 3 divides the rate by 9, the reaction is second-order.