Find ideal $J$ which contain ideal $I$. Consider ring $\Bbb Z[x]$ of polynomials with integer coefficient. 
Let $I = \big\{ f(x) | f\big(\frac{1}{2}\big)=0 \big\}$. 
I have proved that $I$ is prime ideal and $I$ is NOT a maximal ideal. So, find ideal $J$ of $\Bbb Z[x]$ such that, 
$I$ is a proper subset of $J$ and $J$ is a proper subset of $\Bbb Z[x]$. (Such ideal exist because $I$ is not maximal ideal)
 A: Let us show that $I=(2x-1)\mathbb{Z}[X]$.
Working in $\mathbb{Z}[\frac{1}{2}][x]$ and cancelling denominators  shows that for any $g\in\mathbb{Z}[x]$, we may write $2^ng=(2x-1)Q+r$, where $n\geq 0$,$Q\in\mathbb{Z}[x]$,$r\in\mathbb{Z}$. 
In particular, if $g\in I$, we get $r=0$.
If $n=0$, it shows that $g$ is a multiple of $2x-1$.
If $n\geq 1$, reducing modulo $2$ shows that $Q=2S, S\in\mathbb{Z}[X]$. Cancelling by $2$ and using induction shows finally that $g$ is a multiple of $2x-1$. Hence $I=(2x-1)\mathbb{Z}[X]$.
So you just need to find a maximal ideal containing $2x-1$. The ideal $J=(3,2x-1)$ works,  Note that $J$ is indeed maximal. Indeed , apply first isomorphism theorem to $g\in\mathbb{Z}[X]\mapsto \overline{g(2)}\in\mathbb{F}_3$.
In fact, $I$ is contained in infinitely many maximal ideals: if $p>2$ is prime, the ideal $(p, 2x-1)$ works perfectly too.
Side note. How to guess such a $J$. Well, this is easier if you know that the maximal ideals of $\mathbb{Z}[X]$ are the ideals $(p, f)$, where $p$ is prime and $f\in\mathbb{Z}[X]$ , $p$ does not divide the leading coefficient of $f$ and $f$ irreducible modulo $p$.
