See the answer below.
Work Step by Step
From the table, quadrupling [NO], at constant [Br2] results in the initial rate being multiplied by a factor of 16, so it's a reaction of second-order with respect to NO. For Br2, multiplying [Br2] by a factor of 2.5, at a constant [NO], increases the initial rate by a factor of 2.5, so the reaction is of first-order with respect to Br2. Overall, the reaction is of third-order.