You actually only need to know the latitude for that… except the local terrain will play a larger role anyway, unless they started very close to a pole and follow rhumb lines (in this case a meridian/circle of latitude) as opposed to great circles, so better just ask for full coordinates.

We don’t have many data points in the question so let’s extrapolate into the past. There is the hint that they met 8 years earlier at the same spot, during which he’d have gone 1 262 304 000 ft or 384 750.2592 km, completing 9.617 polar circumferences of the Earth (40 007.863 km each).

Huh, that’s not a whole number. In some languages, “eight long years” might mean “a little over 8 years” so let’s assume he finished exactly 10 polar circumnavigations, which took 8 years and about 116 days. Her walked distance over that time is 5x smaller, 2 polar circumnavigations’ worth or 80 015.726 km. This is only exactly 2 great circles if they are polar, but we know that it’s impossible to go due east from either pole. Therefore, she must have gone around a circle of latitude. To end up in the same spot, she must have not-quite-circumnavigated-but-enough-for-Phileas-Fogg the Earth (aka crossed every meridian but not the equator) an integer number of times. After a simple conversion, we can construct a table of the options.

To calcuate latitude from circle-of-latitude circumference (colc), we’ll be using geodetic ↔ ECEF conversion equations (except those with the perverse prime vertical radius of curvature 𝑁 of course) and their notation (simplified with 𝑦 = 0, 𝜆 = 0, ℎ = 0 to ignore longitude and elevation) with values of the WGS-84 ellipsoid. The relationship we’re seeking is between colc/2𝛑 = 𝑝, circle-of-latitude radius, which is at zero longitude equal to ECEF 𝑋, and 𝜙 (latitude). See also Wikipedia on Earth radius by location but remember to skip anything with 𝑁, we’re not doing that.

The geocentric radius (𝑅) is related to 𝜙 (latitude) like this but we only need the distance to axis of rotation 𝑝.

(𝑍/𝑝)(cot 𝜙) = (1 − 𝑒²) → (𝑏²/𝑎²)(𝑍/𝑝) = 1/(cot 𝜙) = tan 𝜙 → 𝜙 = atan((𝑏²/𝑎²)(𝑍/𝑝))
(using 𝑒² = 1 −  𝑏²/𝑎²)

Since sin² 𝛂 = 1 − cos² 𝛂 and we can normalize 𝑍 and 𝑝 to the unit circle with ellipsoid radii 𝑏 and 𝑎 respectively:
𝑍²/𝑏² = (𝑍/𝑏)² = 1 − (𝑝/𝑎)² = 1 − 𝑝²/𝑎², therefore 𝑍 = √(𝑏²(1 − 𝑝²/𝑎²)).

All in all, 𝑝 → 𝜙 conversion is:
𝜙 = atan((𝑏²/𝑎²)(√(𝑏²(1 - 𝑝²/𝑎²))/𝑝))

(Presumably, this could be simpified further but since I can just put this into a calculator so idc)
Per WGS-84:
𝑎 = 6378.137 km
𝑏 = 6352.752 km

Here are the results. Finding appropriate meeting locations at some of the 25+ possible latitudes on either hemisphere is left as an exercise to the reader.

nqcbefPFs	colc = 2𝛑𝑝 [km]	Latitude [°N/°S]
1	too big	N/A
2	6 367.449	3.277975
3	4 244.966	47.934779
4	3 183.724	59.758044
5	2 546.979	66.211738
6	2 122.483	70.346611
7	1 819.271	73.238734
8	1 591.862	75.380740
9	1 414.988	77.033209
10	1 273.489	78.347789
11	1 157.718	79.419029
12	1 061.241	80.309059
13	979.607	81.060439
14	909.635	81.703329
15	848.993	82.259706
16	795.931	82.745975
17	749.111	83.174629
18	707.494	83.555355
19	670.257	83.895779
20	636.744	84.201988
21	606.423	84.478901
22	578.859	84.730536
23	553.691	84.960206
24	530.620	85.170671
25	509.395	85.364245
26	489.803	85.542885