How to calculate the average size of cubic crystals of KCl powder? The total area of $\pu{0.5 g}$ sample of $\ce{KCl}$ powder is measured to be $\pu{1620 cm2}$. If the powder consists of tiny cubic crystals, what is the average size of these crystals?

I'm studying for a physical chemistry final and am not sure how to do this practice question.
I know that the density is 
$$\delta = \frac{m}{A},\label{one}\tag1$$ 
where $m$ is the mass and $A$ is the surface area, so do I just find 
$$A=\frac{m}{\delta}$$
to find the average "size"?
Or do I use \eqref{one} and then find out what the unit cell structure of $\ce{KCl}$ powder is and use that to estimate what the surface area of the crystal is?
For a cube, the surface area is given by the formula $6a^2$ where $a$ is the length of one side. The volume is given by $a^3$. Surface area per volume is 
$$\frac{6a^2}{a^3}= \frac{6}{a}.$$

We know the total surface area to be $\pu{1620 cm^2}$. We can calculate the total volume from the mass of the $\ce{KCl}$ and its density.

\begin{align}
  V &amp;= \frac{m}{\rho}\\
  \rho(\ce{KCl}) &amp;= \pu{1.98 g//cm^3}\\
  V &amp;= \frac{\pu{0.5 g}}{\pu{1.95 g//cm^3}}\\
    &amp;= \pu{0.3 cm^3}
\end{align}

An average crystal should have the same ratio as the total. So we set the two equal and solve for the length.

\begin{align}
  \frac{6}{a} &amp;= \frac{\pu{1650 cm^2}}{\pu{0.3 cm^3}}\\
  a &amp;= \frac{6 \times \pu{0.3 cm^3}}{\pu{1650 cm^2}}\\
  a &amp;= \pu{9E-4 cm}
\end{align}