Let $R$ be a ring and $M$ be a left $R$-module,then I have to show that $ \operatorname{Hom}_R(R,M) \cong M$. But for that I will have to make $ \operatorname{Hom}_R(R,M)$ an $R$-module. I tried $r.\alpha(s)=\alpha(rs)$ as well $r.\alpha(s)=\alpha(sr)$, where $r,s\in R$ and $\alpha \in \operatorname{Hom}_R(R,M)$. But the former didn't qualify for third module axiom and latter didn't make $r.\alpha$ an $R$-linear map.
I know it will be an $R$-module if $R$ is commutative, but it isn't given commutative in the question. Is there an error in question or am I missing something?
The way to make $\text{Hom}_R(R,M)$ into a left $R$-module is to define $$s\cdot f:r\mapsto f(rs).$$ Then $$s'\cdot(s\cdot f):r\mapsto(s\cdot f)(rs')=f(rs's)=((s's)\cdot f)(r)$$ etc.
But the elements of $\text{Hom}_R(R,M)$ all have the form $f_m:r\mapsto rm$ where $m\in M$, and then $m\mapsto f_m$ is the required isomorphism. We check: $$s\cdot f_m:r\mapsto f_m(rs)=rs m=f_{sm}(r).$$