For your use case, where you seem to be expecting extended outages, I would actually recommend trying to get a diesel generator with a small UPS (while you start it up) if possible, or maybe even consider batteries designed for solar power, with the added advantage of saving on future power costs ;)

When you're picking out any kind of power supply, you usually want to look at VA (Volt-Amps), not W (Watts). Nominally, W = V (Volts) * A (Amps), but Watts don't account for "unused" power that's "returned"^{1}. You need to pick a supply that's capable of supplying the peak load you can expect, and a little extra. It also needs to be able to supply an adequate continuous load.

When you want to know how much the electricity is costing you, you use watts. When you are specifying equipment loads, fuses, and wiring sizes you use the VA, or the rms voltage and rms amperage.

That sorts out whether the power supply is capable of supporting your usage. Then you need to look a the capacity, which determines how long the power supply can last with your usage. This is normally measured in Wh (Watt-hours, where 1 WH = 1 Watt for 1 hour) or kWh (kilowatt-hours, 1 kWh = 1000 Wh). To figure out how much you need, you take your continuous load and multiply it by the time you need, with some extra.

Performing the actual calculations with your data:

Your laptop will likely draw 150 W at peak, but far less when idling. I don't have that data, so I'm going to assume a 150 W continuous load. Add in the lights and fan and we reach almost 300 W (rounding and safety). We can usually approximate VA with 1 VA = 0.7 W, though this can vary a lot depending on what the equipment is. 60%-70% is the typical ratio^{2}, and I'm using 70% here. Therefore, we reach about 430 VA. Let's say 450 VA. Remember, this is the value you cannot exceed - so this supply will never support more than about 300 W safely.

At this point you're looking for a power source that can supply at least 450 VA. Next step is to calculate the capacity - 300 W for 6 hours is 300 * 6 = 1600 Wh, or 1.6 kWh. That would require a pretty beefy battery from a UPS. Also note that batteries do degrade over time and will lose capacity and need to be replaced. Again, you might want to add a safety margin, maybe an extra hour. With capacity, if you draw more power it'll just last shorter, and vice versa.

Again, I don't have actual figures for your real continuous load. But you can repeat these calculations with adjusted values.

^{1 See What is the practical difference between watts and VA (volt-amps)?}

^{2 Some loads, including some 80+ certified SMPSes, can almost reach a one-to-one ratio (1 VA > 0.9 W). But it's safer to assume lower than the other way around, unless you can measure and confirm.}