Solution 1
Let the numbers be $x$, $y$, and $z$ in that order. The given tells us that
\begin{eqnarray*}y&=&7z\\ x&=&4(y+z)=4(7z+z)=4(8z)=32z\\ x+y+z&=&32z+7z+z=40z=20\\ z&=&\frac{20}{40}=\frac{1}{2}\\ y&=&7z=7\cdot\frac{1}{2}=\frac{7}{2}\\ x&=&32z=32\cdot\frac{1}{2}=16 \end{eqnarray*}
Therefore, the product of all three numbers is $xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow \boxed{\mathrm{(A)}\ 28}$.

Solution 2
Alternatively, we can set up the system in equation form:
\begin{eqnarray*}1x+1y+1z&=&20\\ 1x-4y-4z&=&0\\ 0x+1y-7z&=&0\\ \end{eqnarray*}
Or, in matrix form
$\begin{bmatrix} 1 & 1 & 1 \\ 1 & -4 & -4 \\ 0 & 1 & -7 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} =\begin{bmatrix} 20 \\ 0 \\ 0 \\ \end{bmatrix}$
To solve this matrix equation, we can rearrange it thus:
$\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -4 & -4 \\ 0 & 1 & -7 \end{bmatrix} ^{-1} \begin{bmatrix} 20 \\ 0 \\ 0 \\ \end{bmatrix}$
Solving this matrix equation by using inverse matrices and matrix multiplication yields
$\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{7}{2} \\ 16 \\ \end{bmatrix}$
Which means that $x = \frac{1}{2}$, $y = \frac{7}{2}$, and $z = 16$. Therefore, $xyz = \frac{1}{2}\cdot\frac{7}{2}\cdot16 = 28 \Rightarrow \boxed{\mathrm{(A)}\ 28}$

Solution 3
Let's denote the 3rd number as $x$, the 2nd as $7x$, and the 1st as $4(7x+x)$ according to the information given in the problem. We know that all three numbers add up to $20$, so $4(8x)+7x+x = 20$. Solving the equation we get $x = \frac{1}{2}$. Then substitute this value of x to solve for the other two numbers. Lastly, we obtain $28$ as the product of all three numbers.