I think I must first prove that I(A) is a subring and then prove that A is in the ideal I(A) and I(A) is the largest sub-ring containing A. but I have problems to show that I(A) is a subring because let x and y in I(A), so $xa\in A$ and $ya \in A$ but I do not know how to prove that $xy\in I(A)$ and the rest.
On
Let $R'$ be a subring of $R$ such that $A\subset R'$ is an ideal, take $x\in R'$ then $\forall a\in A$, $xa\in A$ since $A$ is an ideal in $R'$ so $x\in I(A)$.
Take $x,y\in I(A)$ then $(x+y)a=xa+ya\in A$ for every $a\in A$, so $x+y\in I(A)$. Similarly $x\cdot y \in I(A)$. Also $0_R,1_R\in I(A)$ trivially.
It is easy to see that $A$ is a two sided ideal in $I(A)$, since $A\subset I(A)$ we have that for every $x\in R$ and $a\in A$, $ax\in A$. In particular this happens for every $x\in I(A)$. So $A$ is a right ideal of $I(A)$ and a left ideal of $I(A)$ by construction.
Let $x,y \in I(A)$. Then $xa, ya \in A$ for all $a \in A$. Now $(xy)a = x(ya) = xa'$ for $a' = ya \in A$. Thus $xa' \in A$ and therefore $xy \in I(A)$. That $x+y \in I(A)$ follows from $(x+y)a = xa + xy$ and since $A$ is an ideal. Now since $A$ is a right ideal of $R$ it is also a right ideal in $I(A) \subseteq R$. By definition $A$ is a left ideal in $I(A)$ and thus an ideal. Let now $R' \subseteq R$ be a subring such that $A \subseteq R'$ is an ideal. Then for all $x \in R', a \in A$ we have $xa \in A$. Hence $R' \subseteq I(A)$.