= Solution
An element $\alpha$ is <algebraic element>[algebraic] over $\mathbb Q$ when it is a root of a nonzero polynomial in $\mathbb Q[X]$. Clearing denominators gives
$$
a_0+a_1\alpha+\cdots+a_d\alpha^d=0,
\qquad a_i\in\mathbb Z,quad a_0\ne0.
$$
For $\alpha\ne0$, take $\beta=-(a_1+a_2\alpha+\cdots+a_d\alpha^{d-1})\in\mathbb Z[\alpha]$; then $\alpha\beta=a_0\ne0$.
Since $K=\operatorname{Frac}R$, write $x=a/b$ with $a,b\in R$ and $b\ne0$. Applying the first result to $b$ gives $b\beta=m\in\mathbb Z\setminus\{0\}$, whence $x=a\beta/m$. Changing signs makes $m>0$.
Choose a $\mathbb Z$-basis $e_1,\ldots,e_d$ of $\mathcal O_K$. Write each $e_i=r_i/m_i$ as above and take a common multiple $m$. Then
$$
m\mathcal O_K\subseteq R\subseteq\mathcal O_K.
$$
Thus $R$ is a free abelian group of rank $d=[K:\mathbb Q]$ and has finite index in $\mathcal O_K$. If $0\ne a\in I$, then $aR\subseteq I$ and multiplication by $a$ has nonzero determinant on the rank-$d$ lattice $R$, so $(R:aR)$, and hence $(R:I)$, is finite. Also $m\mathcal O_K\subseteq R$ is an ideal of $R$.
Put $h=(\mathcal O_K:R)$ and $I=m\mathcal O_K$. Then
$$
(R:I)=\frac{m^d}{h},
\qquad
(R:I^2)=\frac{m^{2d}}h.
$$
The assumed multiplicativity of indices gives
$$
\frac{m^{2d}}h=\left(\frac{m^d}h\right)^2,
$$
so $h=1$. Therefore $\boxed{R=\mathcal O_K}$.
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