"For comparison, the US consumes 1.39 x10^9 [eia.gov] litres of fuel per day. According to Wikipedia, the energy density of petrol is 49.2 x 10^6 J/L [wikipedia.org], so that's 684 x10^12 J of energy per day... or, expressed in Watt-days (86400 seconds in a day), that's 7.91 x10^9 W-days of energy."
Wikipedia actually lists 34.2 MJ/L as the energy density of petrol. Since this supports your case, I'll use it.
1.39 x 10^9 L/day * 34.2 x 10^6 J/L = 47.538 x 10^15 J/day. I'm not sure what you did when you calculated daily energy use, but you were off by a couple orders of magnitude.
Converting to watt-days (47.538 x 10^15 / 8.64 * 10^4) gives us 5.502 x 10^11 Watt-days. If we then divide this by 7.68 x 10^12 (20 percent of 6 percent of total sunlight energy falling on arable land, in accordance with your figures), we get about 7.2% of all land needed to meet energy needs, which is a far cry from 1% of all land providing 10 times more energy than we need.
Of course, this is all still a fantasy. Fields need fertilizer or to be planted with crops that will naturally replenish the nitrogen in the soil. If the land isn't 'rested' periodically, yields will drop dramatically. Even with proper farming techniques, yields still will not be close to 100% of the maximum possible biomass. All of this assumes that there is plenty of water to go around; since the majority of US farmland suffered from drought in 2012 (http://www.ers.usda.gov/topics/in-the-news/us-drought-2012-farm-and-food-impacts.aspx), and we have known for a long time that aquifer levels are dropping dangerously low, I'm going to suggest that adequate water is not a safe assumption.
Another consideration is that 7.2% (hopelessly optimistic as it is) refers to the total surface area of ground covered by crops. Even if we planted the crops such that they covered 100% of the planted area at maturity, we still have to consider the full life cycle of the plant from seed to maturity. So, that 6% figure may be correct, but the denominator is much smaller than the field on which the crops are planted.
Also, 4 million square kilometers is way higher than the actual amount of arable land in the United States. You were looking at agricultural land (includes all farmland, including that which is suitable for livestock but not crops). Using your same source, arable land is actually 1,617,800 square km. This adjustment alone would push the 7.2% above to 17.8%, and that is without considering the other factors I listed.
Finally, you have only considered gasoline, when it would be appropriate to include ultra-low sulfur diesel, which is used primarily for transportation. According to (http://www.api.org/~/media/Files/Oil-and-Natural-Gas/Gasoline/US_gasoline-distillate-update.pdf), ULSD production from 2007-2011 is 3.5 million barrels per day. Since the US exports a lot of diesel, and I don't know what percentage of that is actually used in the United States, I'll just split it down the middle and say that half is exported. This translates to 1.038 X 10^16 J/day or 1.201 x 10^11 additional Watt-days. If we count other types of diesel fuel (I don't know other types of diesel fuel are used for, so I just played it safe and assumed they could be replaced by grid power) and assume less than 50% is exported, this number could easily double and would more than triple if we used more recent data and assumed zero exports.
I could keep going, but I think this is sufficient to show that your calculations were off by at least a few orders of magnitude.