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 
$$ \frac{2 E}{F} = 6. $$