Find the order of $\sqrt{3}\in \mathbb{R}$ and $\sqrt{3}\in \mathbb{R}^{*}$.
For the first part, $\mathbb{R}$ is a group under addition. Therefore, $\left \langle \sqrt{3} \right \rangle=\left \{ \sqrt{3},2\sqrt{3},3\sqrt{3},... \right \}$. Therefore, $\sqrt{3}\in \mathbb{R}$ has infinite order.
For the second part, $\mathbb{R}^{*}$ is a group under multiplication. Therefore, $\left \langle \sqrt{3} \right \rangle= \left \{ 3^{1/3},3^{2/3},3,3^{4/3},... \right \}$. Therefore, $\sqrt{3}\in \mathbb{R}^{*}$ also has infinite order.
Is this the correct thinking or am I doing anything wrong?
You are on the right track, but lets make the argument more formal.
The order of an element $g$ from a (multiplicative) group $G$ is a smallest positive integer $n$ such that $g^n=e$ if it exists, otherwise $\infty$.
So in the case of additive $\mathbb{R}$ we are looking at solutions to $n\sqrt{3}=0$. Since $\sqrt{3}>0$ then by induction $n\sqrt{3}>0$ and so this only works for $n=0$. Hence $\sqrt{3}$ has order $\infty$.
In the case of multiplicative $\mathbb{R}^*$ we look at solutions to $\sqrt{3}^n=1$. We can use the fact that $\sqrt{3}>1$ and so (again by induction) $\sqrt{3}^n>1$, which means that the order is $\infty$.