You have 150 W/m^2 hitting your roof each day. You can convert 13% of it into usable energy, and you need 3.5 kW to run your house for a day. Show the MATH,
answer and units, to determine the size solar panel you will need to succeed.

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.