7. A. A uniform cube is marked with the numbers $1,2, \cdots, 6$ on its six faces. If the cube is rolled three times, the probability that the sum of the numbers on the top faces is a multiple of 3 is . $\qquad$
7. A. $\frac{1}{3}$.

Notice that, the sums of the numbers on the top faces of a cube rolled three times that are multiples of 3 are $3, 6, 9, 12, 15, 18$, and
$$
\begin{array}{l}
3=1+1+1, \\
6=1+1+4=1+2+3=2+2+2, \\
9=1+2+6=1+3+5 \\
=1+4+4=2+2+5 \\
=2+3+4=3+3+3, \\
12=1+5+6=2+4+6 \\
=2+5+5=3+3+6 \\
=3+4+5=4+4+4, \\
15=3+6+6=4+5+6=5+5+5, \\
18=6+6+6 .
\end{array}
$$

Let the numbers on the top faces of the cube rolled three times be $x$, $y$, and $z$. Then, among the $(x, y, z)$ that make $x+y+z$ a multiple of 3, there are 8 groups where all three numbers are different, 6 groups where exactly two numbers are the same, and 6 groups where all three numbers are the same.
Thus, the required probability is
$$
\frac{8 \times 3 \times 2+6 \times 3+6}{6 \times 6 \times 6}=\frac{1}{3} \text {. }
$$