Yeah, this works, book is *Topological Graph Theory* by Gross and Tucker, mostly pages 216-217. First we have, on page 4, Theorem 1.1.1 by Euler, the sum of the valences is twice the number of edges. So the average valence is $\frac{2 E}{V}.$

Very much in the same vein, the sum over all the faces is twice the number of edges, so the average edges per face is $\frac{2 E}{F}.$

Your hypothesis is that the average valence is exactly three, so
$$ 3 V = 2 E.   $$

The Euler characteristic says $V - E + F = 0,$ or $F = E - V,$ or 
$$3 F = 3 E - 3 V = 3 E - 2 E = E,$$ so
$6 F = 2 E$ and 
$$ \frac{2 E}{F} = 6. $$ 

There is a figure of $K_7$ on the torus on page 137, the dual would be seven hexagons. So that is the prototype, but of course you can take any large section of hexagon wallpaper and carefully place a rectangle on it to get another example. I don't know what happens if you start with a pentagon and go from there. There is also the one with octagons and squares.