Modular Cubes, Part 2

For a positive number $n$ , define $C (n)$ as the number of the integers $x$ , for which $1 < x < n$ and
$x^{3} \equiv 1 mod n$ .

When $n = 91$ , there are $8$ possible values for $x$ , namely: $9, 16, 22, 29, 53, 74, 79, 81$ .
Thus, $C (91) = 8$ .

Find the sum of the positive numbers $n \leq 10^{11}$ for which $C (n) = 242$ .

Write-up coming later

The complete problem is available here. An approach, code, and answer will be added later.