Solution:

The answer is $\mathbf{(E)}$. To prove it, we want to show that $p(m)$ is even for every integer $m$, and that consequently $p(m)$ is a multiple of 6 if and only if it is a multiple of 3. At this point we can use the hypothesis that the integers $m$ for which $p(m)$ is a multiple of 3 are precisely 40 to obtain the answer.

Let us first observe that the parity of $p(m)$ depends only on the parity of $m$: indeed the parity of a monomial $a \cdot m^{k}$ depends only on the parities of $m$ and of $a$, and the parity of the value of $p(m)$, which is a sum of monomials in $m$, depends only on the parity of each addend. It is therefore sufficient to show that $p(m)$ is even for at least one even value and for at least one odd value of $m$. By hypothesis $p(0)=6$, so it remains to show the existence of an odd value of $m$ for which $p(m)$ is even. We can write $p(x)=x \cdot q(x)+6$, where $q(x)$ is a polynomial with integer coefficients. If $m$ is a multiple of 4, say $m=4k$, then $p(m)=4k \cdot q(4k)+6$ is not a multiple of 4 (it gives remainder 2 upon division by 4). By hypothesis, there are 30 integers between 1 and 60 such that $p(m)$ is a multiple of 4; since the $m$ that are multiples of 4 do not have this property, and the $m$ that are even but not multiples of 4 are only 15, there must exist odd integers such that $p(m)$ is even.