printf epochtime shows wrong values

I suppose that you want to represent the date: October 19th, 2002 00:00:00, which corresponds to the epoch timestamp that you expect: 1034985600.

In such case, you are doing it wrong. Read the manual:

Broken-down time is stored in the structure tm, which is defined in <time.h> as follows:

struct tm {
    int tm_sec;    /* Seconds (0-60) */
    int tm_min;    /* Minutes (0-59) */
    int tm_hour;   /* Hours (0-23) */
    int tm_mday;   /* Day of the month (1-31) */
    int tm_mon;    /* Month (0-11) */
    int tm_year;   /* Year - 1900 */
    int tm_wday;   /* Day of the week (0-6, Sunday = 0) */
    int tm_yday;   /* Day in the year (0-365, 1 Jan = 0) */
    int tm_isdst;  /* Daylight saving time */
};

Your year should be 2002 – 1900 = 102, and your month should be 9, not 10 (months start from 0 = January).

The correct code is:

#include <stdio.h>
#include <time.h>
#include <inttypes.h>

int main(void) {
    struct tm epochtime = {
        .tm_mday = 19,
        .tm_mon = 9,
        .tm_year = 102,
        .tm_hour = 0,
        .tm_min = 0,
        .tm_sec = 0,
        .tm_isdst = -1
    };

    time_t epochdate = mktime(&epochtime);
    if (epochdate == (time_t)(-1)) {
        perror("mktime failed");
        return 1;
    }

    printf("%" PRIuMAX "\n", (uintmax_t)epochdate);
    return 0;
}

Which correctly outputs 1034985600 as you expect.

The problem with your code is most likely that mktime is not able to correctly represent the “wrong” date you provide and returns -1, which you then print as unsigned and becomes a huge nonsensical number.