= Solution
For $\sigma>0$, use the unnormalized <normal density> $h(x)=e^{-x^2/(2\sigma^2)}$. Its ratio region is
$$
0<u\le1,\qquad |v|\le2\sigma u\sqrt{-\log u}.
$$
The bound on $u$ is one because $h$ is maximized at zero. To bound $|v|$, maximize $x\sqrt{h(x)}=xe^{-x^2/(4\sigma^2)}$ for $x\ge0$. Its derivative vanishes at $x=\sqrt2\sigma$, giving $b=\sigma\sqrt{2/e}$. Thus the <normal ratio-of-uniforms envelope> is $0<u<1$, $-b<v<b$.
Draw <independent> $U_1,U_2\sim\operatorname{Unif}(0,1)$, set $u=U_1$ and $v=b(2U_2-1)$, and accept precisely when
$$
u^2\le\exp\left(-\frac{v^2}{2\sigma^2u^2}\right),
$$
returning $X=v/u$. Equivalently compare logarithms to avoid underflow. Since $Z=\sigma\sqrt{2\pi}$, the region area is $\sigma\sqrt{\pi/2}$ and the rectangle area is $2\sigma\sqrt{2/e}$. Therefore
$$
\boxed{r=\frac{\sqrt{\pi e}}4\approx0.7306.}
$$
It is <independent> of the normal scale. In the diagram the vertical coordinate is normalized by $\sigma$, so the same shape works for every positive scale.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-40-ratio-of-uniforms.png]
{title=Normal ratio-of-uniforms region and its minimum axis-aligned rejection rectangle}
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