How to calculate the average size of cubic crystals of KCl powder?
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}