For which $k$ the number $N = 101 ... 0101$ with $k$ ones is a prime?
1. We start with the number \( N \) which is composed of \( k \) ones. We can express \( N \) in terms of powers of 10. Specifically, \( N \) can be written as:
   \[
   N_k = \frac{10^{2k} - 1}{99}
   \]
   This is because \( N \) is a repunit number in base 10, and the formula for a repunit number with \( k \) ones is \( \frac{10^k - 1}{9} \). Since \( N \) has \( k \) ones followed by \( k \) zeros and then another one, it can be expressed as \( \frac{10^{2k} - 1}{99} \).

2. We need to determine for which values of \( k \) the number \( N_k \) is prime. We first consider the case when \( k \) is odd. If \( k \) is odd, we can factorize \( 10^{2k} - 1 \) as follows:
   \[
   10^{2k} - 1 = (10^k - 1)(10^k + 1)
   \]
   Therefore,
   \[
   N_k = \frac{10^{2k} - 1}{99} = \frac{(10^k - 1)(10^k + 1)}{99}
   \]
   Since \( 10^k - 1 \) and \( 10^k + 1 \) are both greater than 1 for \( k > 1 \), \( N_k \) is not prime because it is a product of two numbers greater than 1.

3. Next, we consider the case when \( k \) is even. Let \( k = 2^a l \) where \( a \ge 1 \) and \( l \) is odd. We can write:
   \[
   10^{2k} - 1 = 10^{2^{a+1}l} - 1
   \]
   This can be factored using the difference of squares repeatedly:
   \[
   10^{2^{a+1}l} - 1 = (10^{2^a l} - 1)(10^{2^a l} + 1)
   \]
   Continuing this factorization, we get:
   \[
   10^{2^{a+1}l} - 1 = (10^l - 1)(10^l + 1)(10^{2l} + 1) \cdots (10^{2^a l} + 1)
   \]
   Dividing by 99, we get:
   \[
   N_k = \frac{10^{2k} - 1}{99} = \frac{(10^l - 1)(10^l + 1)(10^{2l} + 1) \cdots (10^{2^a l} + 1)}{99}
   \]
   For \( N_k \) to be prime, the expression must reduce to a single prime factor. This can only happen if \( l = 1 \) and \( a = 1 \), which implies \( k = 2 \).

4. For \( k = 2 \), we have:
   \[
   N = 101
   \]
   We check if 101 is prime. Indeed, 101 is a prime number.

Conclusion:
The only value of \( k \) for which \( N \) is prime is \( k = 2 \).

The final answer is \( \boxed{ k = 2 } \).