It is assumed that the basic facts about the length of the period of a decimal fraction are known. Here is an example how such a period length can actually be computed, even with modest means.
The example is with a prime number, p=523. In this case, it is known that the period length is a factor of p-1. For general (not necessarily prime) n the same method can be employed, but then not with n-1 but with phi(n) which in turn is n-1 if and only if n is prime. In practice, this is not recommended: in order to compute phi(n), one has to decompose n into prime factors, and if one has done that, it is easier to compute the period lengths for the factors of n separately.
The factorisation of p-1 is: 522 = 2 * 3 * 3 * 29
|i||10i mod 523||equals|
|7||107 mod 523||240|
|14||2402 mod 523||70|
|29||(702*10) mod 523||361|
|58||3612 mod 523||94|
|87||(361*94) mod 523||462|
|174||4622 mod 523||60|
|261||(462*60) mod 523||1|
This computation always ends with 1 because ap-1 mod p = 1 for all 0<a<p. If not, then either p is not a prime or you have made a mistake somewhere.
The period length has to be a factor of 261 but none of the numbers in the list. The only factors that remain are 1, 2, 3, 6, 9, 18.
(Since 18 is so small, you can take a shortcut: 109 mod 523 is neither 1 nor -1, so there is no need to compute 1018 mod 523.)
No further i with 10i mod 523 are found. Thus the answer is 261.