Use Python 3:

s, n = input().strip(), int(input().strip())
print(s.count("a") * (n // len(s)) + s[:n % len(s)].count("a"))

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    String s = in.next();
    long n = in.nextLong();
    long count =0;
    for(char c : s.toCharArray())
        if(c == 'a')
        count++;

     long factor = (n/s.length());
     long rem = (n%s.length());
     count =  factor*count  ;
    for(int i=0;i<rem;i++)
        if(s.charAt(i)=='a')
                count++;        
    System.out.println(count);



}

When you get time-outs, you are probably trying to construct the 'infinite' string based on the amount of chars you need. Don't do this, but instead count the number of 'a's in the original string and then multiply this till you get to the target. The remaining part of a string can be calculated then.

Java Solution

static long count(String s, long n) {
    long numOfS = n/s.length();
    long rest = n % s.length();

    if(!s.contains("a")) return 0;
    return s.length()>n? aCounter(s, rest) 
            : numOfS*aCounter(s, s.length()) + aCounter(s, rest);
}

    private static long aCounter(String s, long end) {
        int a=0;
        for (int i = 0; i < end; i++) {
            if (s.charAt(i) == 'a') a++;
        }
        return a;
}

Efficient JavaScript solution

The algorithm doesn't count the string a second time, both best and worst case is O(n) where n = length of the string

From my repo https://github.com/gabrielgiordan/HackerRank

function repeatedString(s, n) {
  let c = 0,
      ca = 0,
      r = n % s.length

  for (let i = s.length; i-- > 0;) {
    if (s.charAt(i) == 'a') {
      ++c

      if (i < r)
        ++ca
    }
  }

  return ((n - r) / s.length * c) + ca
}