Seven children, each with the same birthday, were born in seven consecutive years. The sum of the ages of the youngest three children in $42$. What is the sum of the ages of the oldest three?

$ \mathrm{(A) \ } 51 \qquad \mathrm{(B) \ } 54 \qquad \mathrm {(C) \ } 57 \qquad \mathrm{(D) \ } 60 \qquad \mathrm{(E) \ } 63$
1. Let the age of the youngest child be \( a \).
2. The ages of the youngest three children are \( a \), \( a+1 \), and \( a+2 \).
3. According to the problem, the sum of the ages of the youngest three children is 42:
   \[
   a + (a+1) + (a+2) = 42
   \]
4. Simplify the equation:
   \[
   a + a + 1 + a + 2 = 42
   \]
   \[
   3a + 3 = 42
   \]
5. Solve for \( a \):
   \[
   3a + 3 = 42
   \]
   \[
   3a = 42 - 3
   \]
   \[
   3a = 39
   \]
   \[
   a = 13
   \]
6. The ages of the oldest three children are \( a+4 \), \( a+5 \), and \( a+6 \).
7. Substitute \( a = 13 \) into the expressions for the ages of the oldest three children:
   \[
   (a+4) + (a+5) + (a+6) = (13+4) + (13+5) + (13+6)
   \]
   \[
   = 17 + 18 + 19
   \]
8. Calculate the sum:
   \[
   17 + 18 + 19 = 54
   \]

The final answer is \(\boxed{54}\).