Let $F=\frac{6x^2+16x+3m}{6}$ be the square of an expression which is linear in $x$. Then $m$ has a particular value between:
$\text{(A) } 3 \text{ and } 4\quad \text{(B) } 4 \text{ and } 5\quad \text{(C) } 5 \text{ and } 6\quad \text{(D) } -4 \text{ and } -3\quad \text{(E) } -6 \text{ and } -5$
The quadratic can also be written as
\[x^2 + \frac{8}{3}x + \frac{m}{2}\]
In order for the quadratic to be a square of a linear expression, the constant term must be the square of the x-term divided by 4.  Thus,
\[\frac{m}{2} = (\frac{8}{3})^2 \cdot \frac{1}{4}\]
\[\frac{m}{2} = \frac{16}{9}\]
\[m = \frac{32}{9}\]
The answer is $\boxed{\textbf{(A)}}$.