$\text{Rate of a reaction changes from }2.48\times 10{}^{-3}\text{ mol}{l}^{-1}\text{sec}{}^{-1}\text{to}4.96\times 10{}^{-3}\text{mol}{l}^{-1}\text{sec}{}^{-1}\text{ when concentration of reactant is changed from}\mathrm{}$

$\mathrm{ 0.6}\text{M to}2.4\text{M respectively, the order of reaction is:}$

The correct option is 0.5.

Let's understand how the order of the reaction is determined step-by-step.

Step 1: Understand the Rate Law
The rate of a chemical reaction ($R$) depends on the concentration of the reactant ($\left[A\right]$) raised to the power of its order ($n$). This relationship is given by the rate law formula:
$$R=k{\left[A\right]}^n$$
where:
- $R$ is the rate of reaction,
- $k$ is the rate constant,
- $\left[A\right]$ is the concentration of the reactant, and
- $n$ is the order of the reaction.

Step 2: Set up equations with the given data
From the question, we have two different sets of conditions:
1. Initial rate, $R_1=2.48\times {10}^{-3}{\text{mol L}}^{-1}{\text{s}}^{-1}$ when the concentration ${\left[A\right]}_1=0.6\text{M}$.
This gives the equation:
$$2.48\times {10}^{-3}=k{\left(0.6\right)}^n---\; (Equation\; 1)$$

2. Final rate, $R_2=4.96\times {10}^{-3}{\text{mol L}}^{-1}{\text{s}}^{-1}$ when the concentration ${\left[A\right]}_2=2.4\text{M}$.
This gives the equation:
$$4.96\times {10}^{-3}=k{\left(2.4\right)}^n---\; (Equation\; 2)$$

Step 3: Solve for the order of reaction ($n$)
Divide Equation 2 by Equation 1:
$$\frac{4.96\times {10}^{-3}}{2.48\times {10}^{-3}}=\frac{k{\left(2.4\right)}^n}{k{\left(0.6\right)}^n}$$
Simplifying both sides:
$$2={\left(\frac{2.4}{0.6}\right)}^n$$
Since $\frac{2.4}{0.6}=4$, we have:
$$2=4^n$$
We can write $4$ as $2^2$:
$$2^1={\left(2^2\right)}^n$$
$$2^1=2^{2n}$$

By comparing the exponents:
$$1=2n$$
$$n=\frac{1}{2}=0.5$$

Therefore, the order of the reaction is 0.5.