Let $N$ be an $R$-module. Prove $\mathrm{Hom}_{R-\mathsf{Mod}}(R/I, N) \cong \{n \in N \mid \forall a \in I, an=0\}$.
If $f: R/I \to N$ is an $R$-module homomorphism, then $f(r+I)=rf(1+I)$. Also, since $0=f(I)=f(i+I)=if(1+I)$, then every element $a \in I$ satisfies $an=0$ for all $n \in \mathrm{Im}f$.
Does this tell me something? I am not sure how to proceed.
What this tells you is that a morphism $f : R/I \to N$ is completely determined by $f(1 + I)$. This gives us an idea for a morphism $$ \mathrm{Hom}_R(R/I,N) \to \{n \in N | \forall a \in I, an = 0 \},\; f \mapsto f(1 + I). $$ To show that this map is well defined, show that indeed for all $a \in I$ we have $af(1 + I) = f(a + I) = f(I) = 0$.
Injectivity follows from a function being determined by its image of $1 + I$.
Surjectivity follows from defining for any $n \in N$ such that $an = 0$ the function $f_n$ such that $f_n(1 + I) = n$ and showing that it is well-defined.