A student wishes to determine the thickness of a rectangular piece of aluminum foil but cannot measure it
directly.  She can measure its density $(d)$, length $(l)$, mass $(m)$ and width $(w)$. Which relationship will give the thickness?

$\textbf{(A)}\hspace{.05in}\frac{m}{d \cdot l \cdot w} \qquad\textbf{(B)}\hspace{.05in}\frac{m \cdot l \cdot w}{d} \qquad\textbf{(C)}\hspace{.05in}\frac{d \cdot l \cdot w}{m} \qquad\textbf{(D)}\hspace{.05in}\frac{d \cdot m }{l \cdot w} \qquad$
1. Let:
   - $d$ be the density of the aluminum foil,
   - $l$ be the length of the aluminum foil,
   - $m$ be the mass of the aluminum foil,
   - $w$ be the width of the aluminum foil,
   - $v$ be the volume of the aluminum foil,
   - $t$ be the thickness of the aluminum foil.

2. The volume $v$ of the rectangular piece of aluminum foil can be expressed as:
   \[
   v = l \cdot w \cdot t
   \]

3. The density $d$ is defined as the mass per unit volume:
   \[
   d = \frac{m}{v}
   \]

4. Substituting the expression for volume $v$ into the density formula:
   \[
   d = \frac{m}{l \cdot w \cdot t}
   \]

5. Solving for the thickness $t$:
   \[
   d = \frac{m}{l \cdot w \cdot t} \implies d \cdot l \cdot w \cdot t = m \implies t = \frac{m}{d \cdot l \cdot w}
   \]

6. Therefore, the relationship that gives the thickness $t$ is:
   \[
   t = \frac{m}{d \cdot l \cdot w}
   \]

The final answer is $\boxed{\frac{m}{d \cdot l \cdot w}}$