A cuboid has an integer volume. Three of the faces have different areas, namely $7, 27$, and $L$. What is the smallest possible integer value for $L$?
1. Let the sides of the cuboid be \(a\), \(b\), and \(c\). The volume of the cuboid is given by \(abc\), and the areas of the three different faces are \(ab\), \(ac\), and \(bc\).
2. We are given that the areas of three faces are \(7\), \(27\), and \(L\). Therefore, we have:
   \[
   ab = 7, \quad ac = 27, \quad bc = L
   \]
3. To find the volume \(abc\), we multiply the areas of the three faces and take the square root:
   \[
   \sqrt{(ab)(ac)(bc)} = \sqrt{(abc)^2} = abc
   \]
4. Substituting the given values, we get:
   \[
   \sqrt{7 \cdot 27 \cdot L} = abc
   \]
5. Simplifying inside the square root:
   \[
   \sqrt{189L} = abc
   \]
6. For \(abc\) to be an integer, \(\sqrt{189L}\) must be an integer. This implies that \(189L\) must be a perfect square.
7. The prime factorization of \(189\) is:
   \[
   189 = 3^3 \cdot 7
   \]
8. For \(189L\) to be a perfect square, \(L\) must include the factors necessary to make all exponents in the prime factorization even. Therefore, \(L\) must include at least one factor of \(3\) and one factor of \(7\) to balance the exponents:
   \[
   189L = 3^3 \cdot 7 \cdot L
   \]
9. The smallest \(L\) that makes \(189L\) a perfect square is when \(L\) includes one additional factor of \(3\) and one additional factor of \(7\):
   \[
   L = 3 \cdot 7 = 21
   \]
10. Therefore, the smallest possible integer value for \(L\) is \(21\).

The final answer is \(\boxed{21}\).