Step 1 - Write the rate law.
The reaction takes place in ONE STEP, so it is an ELEMENTARY reaction. For an elementary
reaction, order = molecularity, and the exponents in the rate law are simply the
stoichiometric coefficients:
$$\text{Rate} = k[A]^2[B]^1$$
Overall order $= 2 + 1 = 3$

Step 2 - Let the initial rate be $r_1$.
$$r_1 = k[A]^2[B]$$

Step 3 - Find the new concentrations.
Concentration $=\dfrac{\text{number of moles}}{\text{volume}}$. The number of moles does
not change; only the volume changes.
If volume becomes $\dfrac{V}{3}$, then each concentration becomes THREE TIMES larger:
$$[A]' = 3[A] \qquad [B]' = 3[B]$$

Step 4 - Calculate the new rate $r_2$.
$$r_2 = k(3[A])^2(3[B]) = k \times 9[A]^2 \times 3[B] = 27\,k[A]^2[B]$$
$$\boxed{r_2 = 27\,r_1}$$
The rate of the reaction becomes 27 TIMES the original rate.

Step 5 - Effect on the order of reaction.
The ORDER of a reaction is fixed by the RATE LAW (the mechanism), i.e. by the exponents
of the concentration terms. Changing the volume changes only the VALUES of the
concentrations, not the exponents.
Therefore there is NO CHANGE in the order of reaction - it remains THIRD ORDER.