Here is the number list from the text file:

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Numbers=[117, 25, 199, 149, 183, 50, 114, 150, 149, 22, 13, 26, 86, 15, 151, 147, 16, 110, 16, 75, 65, 60, 114, 143,
         124, 89, 27, 147, 117, 146, 145, 67, 133, 143, 99, 172, 29, 64, 114, 9, 159, 85, 1, 120, 20, 4, 63, 54, 32, 90,
         36, 128, 137, 20, 150, 124, 55, 6, 42, 14, 93, 36, 1, 18, 185, 27, 89, 97, 96, 169, 54, 64, 36, 103, 196, 170, 112,
         191, 87, 179, 150, 50, 197, 167, 166, 190, 13, 116, 109, 35, 7, 61, 3, 111, 163, 11, 87, 113, 12, 174, 29, 162, 7, 10,
         190, 65, 97, 155, 152, 166, 199, 139, 76, 48, 178, 6, 164, 54, 11, 143, 134, 21, 198, 70, 107, 78, 5, 131, 62, 112, 155,
         122, 35, 94, 59, 142, 78, 108, 198, 156, 184, 98, 176, 36, 165, 154, 75, 80, 142, 26, 110, 75, 108, 134, 7, 57, 175, 76, 191,
         8, 123, 124, 53, 167, 76, 48, 81, 35, 132, 27, 160, 185, 82, 87, 145, 13, 154, 138, 3, 72, 75, 104, 71, 168, 15, 149, 61, 21,
         90, 103, 108, 39, 104, 151, 158, 75, 10, 142, 13, 130]

And here is the code that finds the required number. I think it is fairly easy to understand

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for Number in Numbers:
    if Numbers.count(i)==1:
        print Number
        break

In pascal:

$\boxed{\begin{array}{|} \text{uses crt;}\\ \text{var a:array[1..200]of integer;}\\ \text{f:text;}\\ \text{const fi='TEST.TXT';}\\ \text{procedure input;}\\ \text{var i:byte;}\\ \text{begin}\\ \text{assign(f,fi);}\\ \text{reset(f);}\\ \text{for i:=1 to 200 do readln(f,a[i]);}\\ \text{end;}\\ \text{procedure output;}\\ \text{var i,v:byte;}\\ \text{begin}\\ \text{for i:=1 to 199 do begin}\\ \text{v:=i+1;}\\ \text{while a[i]<>a[v] do begin}\\ \text{v:=v+1;}\\ \text{if v=200 then begin writeln(a[i]);exit;end;}\\ \text{end;}\\ \text{end;}\\ \text{end;}\\ \text{begin}\\ \text{clrscr;}\\ \text{input;}\\ \text{output;}\\ \text{readln;}\\ \text{end.}\end{array}}$

Show the answer: 25.

CTRL+F on any browser and search number by number. It's pretty easy because the answer is the second appearance on the list...

C++

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#include <stdio.h>
#include <iostream>
#include <vector>
using namespace std;

int numAppearance(vector <int> v, int n){
    int num = 0;
    for (int i = 0; i < v.size(); i++){
        if (v[i] == n)
            num++;
    }

    return num;
}

int nonrepeat(vector<int> v){
    for (int i = 0; i < v.size(); i++){
        if (numAppearance(v, v[i]) == 1)
            return v[i];
    }
}

int main(){

    freopen("answer.in", "r", stdin);
    freopen("answer.out", "w", stdout);

    int n = 200;

    vector <int> v(n);
    for (int i = 0; i < n; i++){
        cin >> v[i];
    }

    cout << nonrepeat(v);

    return 0;
}

Although I'm not averse to using computer time rather than my own time for solving problems I would like to point out that using count on every element of the list uses computer time unnecessarily. Better to count the number of occurences of each item in the list first. There's a class in the standard Python collections module that makes this easy to do.