Let $n$ be an integer.  If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?
1. **Identify the possible units digits of perfect squares:**
   The units digit of a perfect square can only be one of the following: \(0, 1, 4, 5, 6, 9\). This is because:
   - \(0^2 = 0\)
   - \(1^2 = 1\)
   - \(2^2 = 4\)
   - \(3^2 = 9\)
   - \(4^2 = 16\) (units digit is 6)
   - \(5^2 = 25\) (units digit is 5)
   - \(6^2 = 36\) (units digit is 6)
   - \(7^2 = 49\) (units digit is 9)
   - \(8^2 = 64\) (units digit is 4)
   - \(9^2 = 81\) (units digit is 1)

2. **Express \(n^2\) in terms of its digits:**
   Let \(n^2 = 100a + 10b + c\), where \(a\) is the hundreds digit, \(b\) is the tens digit, and \(c\) is the units digit. Given that the tens digit of \(n^2\) is 7, we have:
   \[
   100a + 70 + c
   \]
   This implies that \(10b = 70\), so \(b = 7\).

3. **Consider the congruence modulo 4:**
   Since \(n^2 \equiv 0, 1 \pmod{4}\), we need to check the possible values of \(c\) that satisfy this condition. We have:
   \[
   100a + 70 + c \equiv 2 + c \pmod{4}
   \]
   Therefore, we need:
   \[
   2 + c \equiv 0 \pmod{4} \quad \text{or} \quad 2 + c \equiv 1 \pmod{4}
   \]
   This simplifies to:
   \[
   c \equiv 2 \pmod{4} \quad \text{or} \quad c \equiv 3 \pmod{4}
   \]

4. **Check the possible units digits:**
   The possible units digits of perfect squares are \(0, 1, 4, 5, 6, 9\). We need to find which of these satisfy \(c \equiv 2 \pmod{4}\) or \(c \equiv 3 \pmod{4}\):
   - \(0 \equiv 0 \pmod{4}\)
   - \(1 \equiv 1 \pmod{4}\)
   - \(4 \equiv 0 \pmod{4}\)
   - \(5 \equiv 1 \pmod{4}\)
   - \(6 \equiv 2 \pmod{4}\)
   - \(9 \equiv 1 \pmod{4}\)

   The only value that satisfies \(c \equiv 2 \pmod{4}\) is \(6\).

Conclusion:
The units digit of \(n^2\) is \(\boxed{6}\).