Defintion: an element $a\in R$ is said to be right regular if, there is no nonzero element $b\in R$ such that $ab=0$.
My Questuion: Let $R$ be a ring with identity. Then $R$ is $R$-isomorphic to a right ideal $I$ of $R$ (as a right $R$-module) if and only if $I=aR$ for some right regular element of $R$.
Now I have find a solution of my problem which is as follows:
First suppose that there exists an $R$-isomorphism $f:R\rightarrow I$. Let $a=f(1)\in R$. Then $ab=0\implies f(1)b=0\implies f(b)=0\implies b=0$ for any $b\in R$, as $f$ is one-one. Let $s\in I$ then there exist $r\in R$ such that $s=f(r)=f(1)r=ar$, as $f$ is onto. Thus $a$ is a right regular element such that $I=aR$.
Conversely suppose that there exists a right regular element $a\in R$ such that $I=aR$. Define a map $f:R\rightarrow I$ by $f(r)=ar,\forall r\in R$. Then $f$ is an $R$-isomorphism.