A topology on $R$ defined by the filtration $\{I_n\}$ is Hausdorff if and only if the intersection of all $I_n$ is the zero ideal (0). A topology on $R$ defined by the  filtration $\{I_n\}$ is Hausdorff  if and only if the intersection of all $I_n$ is the zero ideal (0).

I tried but I could not get. Any help will be highly appreciated
The ideals in the filtration form a basis of neighborhoods of zero.

In general topological spaces, we have that $\overline{\{x\}}$ (the closure of the singleton $\{x\}$) is the intersection of the neighborhoods of $x$ or, as well, the intersection of the neighborhoods in a basis.

Since in a Hausdorff space singletons are closed, one of the implications is clear: if $R$ is Hausdorff, then the intersections of the filtration ideals is $\{0\}$.

For the converse implication, you need to know that $\{0\}$ being closed is equivalent to the topological ring $R$ being Hausdorff. This is even easier when the topology is given by a filtration. If $x\ne y$, then $x-y\ne0$, so there exists $n$ such that $x-y\notin I_n$. Then $(x+I_n)\cap(y+I_n)=\emptyset$: indeed, if $z\in(x+I_n)\cap(y+I_n)$ we have
$$
z=x+a=y+b
$$
with $a,b\in I_n$, so $x-y=b-a\in I_n$ contrary to the hypothesis.
Here you can find the full hint: