In general $Ext^1_R(R/I,M) \not\cong M/IM$.
As an example take a finite abelian group $G$ and set $R := \mathbb{Z}G$ and let $I:= I_G = \ker(\mathbb{Z}G \to \mathbb{Z},\; g \mapsto 1)$ be the augmentation ideal. $I_G$ is a prime ideal since $\mathbb{Z}G/I_G \cong \mathbb{Z}$. Then, with trivial coefficients $$Ext_R^1(R/I,\mathbb{Z})=Ext_{ZG}^1(\mathbb{Z},\mathbb{Z})=H^1(G;\mathbb{Z})=Hom(G,\mathbb{Z})=0$$ while $\mathbb{Z}/I_G\mathbb{Z}=\mathbb{Z}$. The latter holds because $I_G$ is generated by $g-1$ $(g \in G)$ and $(g-1) \cdot 1 = 0$.
In general, the following holds: $$Ext^1_R(R/I,M)=\dfrac{Hom_R(I,M)}{i(M)}$$ where $i: M \to Hom_R(I,M),\; m \mapsto (x \mapsto xm)$. But it seems hard to get a closed expression for $Hom_R(I,M)$ if
If $I$ is not free asfinitely generated by $R$-module$a_1,...,a_n$ we can choose a presentation $R^n \to I \to 0$ that induces an embedding $0 \to Hom_R(I,M) \to Hom_R(R^n,M) \cong M^n$. Thus $$Ext_R^1(R/I,M) \le \dfrac{M^n}{\lbrace (a_1m,...,a_nm) \mid m \in M \rbrace}.$$