Why is my 2nd for loop(the one with j) is not working in this C program?

In this case strlen() is the wrong option. As Rick points out, strlen() expects a C style character string. Now you could, of course, alter custIDs variable to add a nul char at the end, e.g.

char custIDs[11] = { 'A', 'B', 'D', 'E', 'F', 'G', 'H' 'I', 'J', 0};

but that just abuses the idea of what a custID might be and how C strings work. A better solution would be to use a #define to capture the number of customers in the "database" e.g.

#include <stdio.h>

#define CUST_SIZE 10

int main()
{
    int i, j;
    char custIDs[CUST_SIZE] = { 'A' ... 'J' };
    /* other definitions, etc ... */
      

    for(i = 0; i < CUST_SIZE; ++i)
    {
       /* do work */
    }

    for( j = 0; j < CUST_SIZE; ++j )
    {
       /* do other work */
    }
}

Note that we really don't need separate variables for each for loop, since the two loops are independent of each other. Better yet, if you have C99 or later (confusingly, perhaps, that's C11) you an use a for looped scoped variable, e.g.

for(size_t i = 0; i < CUST_SIZE < ++i)

In C, using an array of chars as a string means that the string is 0 terminated, the last char must be a 0. As already told in previous answers.

By the way, what happens if a customer have a balance of exactly 100 ?

Quote:

Why is my 2nd for loop(the one with j) is not working in this C program?

Your code do not behave the way you expect, or you don't understand why !

There is an almost universal solution: Run your code on debugger step by step, inspect variables.
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't know what your code is supposed to do, it don't find bugs, it just help you to by showing you what is going on. When the code don't do what is expected, you are close to a bug.
To see what your code is doing: Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute.

Debugger - Wikipedia, the free encyclopedia[^]

Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]

1.11 — Debugging your program (stepping and breakpoints) | Learn C++[^]

The debugger is here to only show you what your code is doing and your task is to compare with what it should do.

In addition to all the other suggestions, you only need a single loop:

for( i = 0 ; i< CUST_SIZE ; i++) // see solution by k5054
{
    if(custbalances[i] > 100 )
    {
        printf(" %c is a defaulter!\n", custIDs[i]);
    }
    else
    {
        printf(" %c is all good\n ", custIDs[j]);
    }
}