Alright, so in the reaction
$$\ce{2NOBr(g) -> 2NO(g) + Br2}$$
the rate law is
$$\text{rate} = k[\ce{NOBr}]^x$$
where $k$ is some constant and $x$ is the order of the reaction in respect to $\ce{NOBr}$.
By seeing how the initial rate changes when we change the concentration of $\ce{NOBr}$, we can determine the value of $x$. We can use any two of the three. I'm going to use the first and third trials. If we divide them we get
$$\frac{\text{rate 3}}{\text{rate 1}} = \frac{k[\ce{NOBr}]_3^x}{k[\ce{NOBr}]_1^x}$$
$$\frac{2.43\times10^2}{1.08\times10^2} = \frac{0.900^x}{0.600^x}$$
The $k$'s cancel out.
$$2.25 = 1.5^x$$
$$x = 2$$
The rate is second order in respect to $\ce{NOBr}$, and the rate law is written $\text{rate} = k[\ce{NOBr}]^2$. If you double the concentration, the rate will quadruple.
$$\text{rate before doubling concentration} = k[\ce{NOBr}]^2$$
$$\begin{align}\text{rate after concentration} &= \left(2[\ce{NOBr}]\right)^2\\
&= 2^2[\ce{NOBr}]^2\\
&= 4[\ce{NOBr}]^2\\
&= 4 \times \text{rate before doubling concentration}\end{align}$$
A tripling of the concentration will increase the rate by a factor of nine, a quadrupling of the concentration increases the rate by a factor of 16, and so on.