Field of fractions of $\mathbb Z[X]/(2X-1)$

${\mathbb Z}[X]/(2X - 1) \cong {\mathbb Z}[\frac{1}{2}]$, so, indeed, its field of fractions is ${\mathbb Q}$.

(And, by the way, $(2X - 1)$ is not a maximal ideal: ${\mathbb Z}[\frac{1}{2}$] is not a field.)

The unique ring homomorphism $\varphi:\ \Bbb{Z}[X]\ \longrightarrow\ \Bbb{Q}$ mapping $X$ to $\frac{1}{2}$ has kernel $\ker\varphi=(2X-1)$ yielding an injection $\Bbb{Z}[X]/(2X-1)\ \hookrightarrow\ \Bbb{Q}$, so the field of fractions of $\Bbb{Z}[X]/(2X-1)$ is contained in $\Bbb{Q}$. Of course the only subfield of $\Bbb{Q}$ is $\Bbb{Q}$ itself, so this is indeed the field of fractions of $\Bbb{Z}[X]/(2X-1)$.