To determine the nature of the decimal expansion of \(\sqrt{27}\), we first simplify the expression:
\[ \sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3 \sqrt{3} \]
The square root of 3, \(\sqrt{3}\), is an irrational number, which means it cannot be expressed as a fraction of two integers and has a non-repeating, non-terminating decimal expansion.
Since \(\sqrt{27} = 3\sqrt{3}\), and \(\sqrt{3}\) is irrational, it follows that \(\sqrt{27}\) also has a non-terminating, non-repeating decimal expansion.
Therefore, the correct answer is:
B. a non-terminating, non-repeating decimal.