why should i declare 128*1024 size of memory for 256KB described in the problem?
my code goes here:
#include<iostream>
#include <cstdio>
#include <cmath>

double buf[128 * 1024];
int main()
{
    int i = 0;
    unsigned long long n;
    while (scanf("%llu", &n) != EOF) {
        buf[i ++] = (double)sqrt(n * 1.0);
    }
    for (; --i >= 0; ) {
        printf("%.4lf\n", buf[i]);
    }
    return 0;
}

//  why 128*1024  ?

Because 256kb means there's at most 128kb of digits and 128kb of spaces?

Here the array size is 128*1024 and it is double. So doesn't the array hold 128*1024*8 KB of memory(8 bytes for each element) ?

Well you don't necessarily need double, you may use long long int for keeping numbers.
Of course, it's 8 bytes either way, but you just don't know whether your numbers are going to be a few huge ones, or a ton of 1-digit ones, so you just have to be on a safe side.