Instead of carrying the mass through the chain one step at a time, we can combine all four percentage factors into a single overall conversion fraction first, and then apply it to the 1 kg of OFMSW in one multiplication. Each kilogram of OFMSW passes through four successive reductions before it becomes methane-generating BVS: it loses moisture, retaining a dry-solids fraction of $f_1 = 1 - 0.20 = 0.80$; of that, the volatile fraction is $f_2 = 0.90$; of the volatile solids, the biodegradable fraction is $f_3 = 0.70$; and of the biodegradable solids, the fraction actually converted is $f_4 = 0.85$. The combined fraction of the original 1 kg of OFMSW that ends up as converted BVS is
\[ f = f_1 \times f_2 \times f_3 \times f_4 = 0.80 \times 0.90 \times 0.70 \times 0.85 \]Multiplying these out step by step, $0.80 \times 0.90 = 0.72$, then $0.72 \times 0.70 = 0.504$, then $0.504 \times 0.85 = 0.4284$. So exactly $0.4284$ kg out of every 1 kg of OFMSW ends up as BVS that is converted during digestion. Since each kilogram of converted BVS yields 12 m3 of methane gas, the total gas volume from 1 kg of OFMSW is
\[ V_{gas} = f \times 12 = 0.4284 \times 12 = 5.1408\ \text{m}^3 \]Rounding to two decimal places,
\[\boxed{V_{gas} = 5.14\ \text{m}^3}\]