Answer
$q=3.54\mu C$
Work Step by Step
We know that:
$\phi=\frac{q}{\epsilon_{\circ}}$
This can be rearranged and rewritten as:
$q=\epsilon_{\circ}\phi$
$\implies q=\epsilon_{\circ}A({E_l-E_u})$ where $E_u$ and $E_l$ are the magnitudes of electric fields at upper and lower ends of the cube respectively.
Therefore;
$q=8.85\times 10^{-12}(100)^2(100-60.0)=3.54\times 10^{-6}=3.54\mu C$
