A positive integer's, n, reciprocal, $\frac{1}{n}$, in which the decimal's repetend has a length of three digits which starts at the decimal mark.
e.g. 0.037037... of the integer, 27 ,reciprocal $\frac{1}{27}$
where p is the 3-digit-repetend ( e.g. according to the previous example $p=37$.
27 and 37 are such integers, but how are the rest deduced?
On
$$\frac 1n=0.abcabcabc......\Rightarrow\frac 1n = \frac{abc}{10^3}(1+\frac{1}{10^3}+....+\frac{1}{10^{3n}}+....)=\frac{abc}{999}$$ It follows $$abc=\frac{999}{n}=\frac{3^3\cdot 37}{n}$$ $3^3\cdot 37$ has $(3+1)(1+1)=8$ factors from which we have eight solutions for $n$. The "trivial" ones are $1,3,9$. The other five are $27,37,111,333,999$ whose decimal representations have periods $037,027,009,003$ and $001$ respectively.
(I called "trivial" $1,3,9$ because $\frac 11=0.99999999....;\frac 13= 0.333333.....;\frac 19= 0.11111....$)
It should at least be a factor of $999$. So,
$$999 = 3^3 \cdot 37$$
Your choices then are $1, 3, 9, 27, 37, 111, 333,$ and $999$. The choices $n=1, 3,$ and $9$ have repetends of length $1$. All others have repetend of length $3$.
As for why this works, let's say that $k = 0.abcabcabc ...$ is a repeating decimal with a pattern of three digits $a,b,c$ (not necessarily different digits). Then, $1000k = abc.abcabcabc ...$, and subtracting the two, $999k = abc$, so $k = abc/999$. Then, to get $k$ into a form $1/n$, $abc$ must divide $999$.