A rectangular box measures $a\times b\times c$, where $a$, $b$, and $c$ are integers and $1\le a\le b\le c$. The volume of the box is twice its surface area. How many ordered triples $(a,b,c)$ are possible?

1. $abc=2\cdot2(ab+bc+ca)$ gives $\frac1a+\frac1b+\frac1c=\frac14$. Since $a\le b\le c$, $\frac3a\ge\frac14$, so $a\le12$; also $a>4$.
2. For each $a$ solve $\frac1b+\frac1c=\frac14-\frac1a$ with $a\le b\le c$ by clearing denominators into a product of two factors equal to a constant:
   \begin{itemize}
3. $a=5$: $(b-20)(c-20)=400$, with $b-20\in\{1,2,4,5,8,10,16,20\}$: $8$ solutions.
4. $a=6$: $(b-12)(c-12)=144$, with $b-12\in\{1,2,3,4,6,8,9,12\}$: $8$ solutions.
5. $a=7$: $(3b-28)(3c-28)=784$, with $3b-28\in\{2,8,14\}$ (the factor must be $\equiv2\pmod3$ and give $b\ge7$): $b=10,12,14$, $3$ solutions.
6. $a=8$: $(b-8)(c-8)=64$, with $b-8\in\{1,2,4,8\}$: $4$ solutions.
7. $a=9$: $(5b-36)(5c-36)=1296$, with $5b-36\in\{9,24\}$ and $b\ge9$: $b=9,12$, $2$ solutions.
8. $a=10$: $(3b-20)(3c-20)=400$, with $3b-20\in\{10,16\}$ and $b\ge10$: $b=10,12$, $2$ solutions.
9. $a=11$: $(7b-44)(7c-44)=1936$ has no solution with $b\ge11$.
10. $a=12$: $b=c=12$, $1$ solution.
    \end{itemize}
11. The total is $8+8+3+4+2+2+0+1=\boxed{28}$.