If a whole number has exactly three factors, it is a perfect square number, and its square root is prime. In the case of a number like $m$, the only three factors will be $\(1, \sqrt{m}\)$ (which itself has no further factors), and the number itself.
Let's examine the answer choices:
(A) $m$ is a prime number

Prime numbers have exactly two factors: 1 and themselves. They never have three factors. This is not true.
(B) $m$ is the square of another number

This must be true; otherwise, $m$ would have an even number of factors. The only way to have an odd number of factors is for two of the factors to be the same-and this only happens in square numbers.
(C) $m$ is a composite number

A composite number is a number that is not prime. Since we know that $m$ is a square number, it cannot be prime, so it is indeed composite. True.
(D) $\(\sqrt{m}\)$ is a prime number

This must be true. We know that $m$ has an integral square root, but if that square root were not prime then it itself would be factorable, and those factors would also be factors of $m$-yielding a total of more than three factors.
(E) $m$ is the cube of another number

This cannot be true: if $m$ were the cube of another number-let's call it $\(x\)$-then $\(x^2\)$ would be a factor of $m$ as well, and there would be at least four total factors of $m$.

(F) $m$ is odd

This might seem to be always true, now that we've established that $m$ is the square of a prime whole number. We also know that the square of an odd number is also an odd number; the only square numbers that are even are the squares of even numbers. So are there any prime numbers that are not odd? Yes! 2 is the only even prime number, which means that $m$ could equal 4. Because it's possible that $m=4$, even if all other possible values of $m$ are odd, we cannot categorically state that $m$ is odd.

The correct answer choices are (B), (C), and (D).