Answer:
Energy = .13 W / m^2 energy of incident energy
N = 3500 Watts / day power needed
N = 3500 Watts (3600 * 24 sec) = .0405 Watts/sec
The problem must mean that one needs 3.5 Kw-days
3.5 Kw-days = 3500 watts * 86400 sec = 3.02E8 joules
150 J/sec-m^2 * .13 = 19.5 J / sec-m^2 usable energy
In one day 19.5 J/sec-m^2 = 1.68E6 J/m^2 usable energy received
Area = 3.028E8 J / 1.68E6 J/m2 = 180 m^2
One would need 180 m^2 of solar panels
That's quite a lot of energy
A 1100 watt microwave oven uses 1.1 kW while running so 3.5 kW for 24 hours seems to be quite a lot.
