There exists $\alpha \in\left[\dfrac{16}{3},\dfrac{17}{3}\right]$ with the required property. To see this, we will construct an interval sequence
$$\left[\dfrac{16}{3},\dfrac{17}{3}\right]=[\alpha_{1},\beta_{1}]\supset [\alpha_{2},\beta_{2}]\supset\cdots\supset[\alpha_{n},\beta_{n}],$$
where $\alpha_{n}$ and $\beta_{n}$ are such that
$$\alpha^n_{n}-\dfrac{1}{3}=\beta^n_{n}-\dfrac{2}{3}=m_{n}\in \Bbb N^{+},$$
so that, for any $x\in [\alpha_{n},\beta_{n}]$, we have
$$\dfrac{1}{3}\le\{x^n\}\le\dfrac{2}{3}.$$
We construct the interval sequence by induction. Assume that we have $[\alpha_{n},\beta_{n}]$. Let
$$a=\alpha^{n+1}_{n},\quad\quad b=\beta^{n+1}_{n}.$$It follows that $$
b-a=(m_{n}+\dfrac{2}{3})\beta_{n}-(m_{n}+\dfrac{1}{3})\alpha_{n}>\dfrac{\alpha_{n}}{3}>\dfrac{5}{3}.$$
Then there exists $m_{n+1}\in \Bbb N^{+}$ such that
$$\left[m_{n+1}+\dfrac{1}{3},m_{n+1}+\dfrac{2}{3}\right]\subset[a,b].$$
We take
$$\alpha_{n+1}=\sqrt[n+1]{m_{n+1}+\dfrac{1}{3}},\qquad\beta_{n+1}=\sqrt[n+1]{m_{n+1}+\frac{2}{3}}.$$
Now $$\alpha^{n+1}_{n}=a<\alpha^{n+1}_{n+1}=m_{n+1}+\dfrac{1}{3}<\beta^{n+1}_{n+1}=m_{n+1}+\dfrac{2}{3}<b=\beta^{n+1}_{n},$$and hence $\alpha_{n}\le\alpha_{n+1}<\beta_{n+1}<\beta_{n},$ or
$$[\alpha_{n},\beta_{n}]\supset[\alpha_{n+1},\beta_{n+1}].$$