Remember that the quotient map induces a lattice isomorphism between the ideals of $R/I$ and the ideals of $R$ that contain $I$; in particular, since $J\cap K$ is the largest ideal of $R$ that contains $I$ and is contained in both $J$ and $K$, then it follows that $(J\cap K)/I$ is the largest ideal of $R/I$ that is contained in both $J/I$ and $K/I$, hence must equal $(J/I) \cap (K/I)$.
Though I think that is by far the best way to go, to prove it explicitly, note that if $x\in J\cap K$, then $x+I\in J/I$ and $x+I\in K/I$, so you get $(J\cap K)/I\subseteq (J/I)\cap(K/I)$. For the converse inclusion, let $a+I\in (J/I)\cap(K/I)$. Then there exists $j\in J$ and $k\in K$ with $j+I = k+I = a+I$. In particular, $j-k\in I$. Therefore, $k=j-(j-k)\in J$, because $j\in J$ and $j-k\in I\subseteq J$, hence $k\in J\cap K$, thus, $a+I = k+I\in (J\cap K)/I$.