The rate of evaporation depends critically on the rate at which water vapor is removed from the water/air interface.
If your cup was closed, water would evaporate until you reach the saturated vapor pressure of water at the prevailing temperature, and then things would reach equilibrium - water vapor would go into the liquid as fast as it evaporated. The fact that water is evaporating at a constant rate tells us that a certain amount of moisture is escaping. Now there are two mechanisms for this: diffusion, and convection. If the air in the vicinity of the cup is perfectly still (imagine a very tall, narrow cup with just a little bit of water in the bottom) then you can compute the diffusion of water through the gradient (almost 100% at the interface, and equal to the relative humidity of the air at the opening). Now you are losing 0.1 mg of water per second (rounding) ; the mass diffusivity of water in air is 0.28 cm$^2$/s. For a cup 10 cm diameter and 5 cm deep, with a 50% RH at the opening, we compute the diffusion as follows:
Saturated vapor pressure at 20C (initial assumption... Let's see if we are in the ball park) is 2.34 kPa so the pressure difference across the gradient is half that - 1.67 kPa. That means the density difference is about 1/60th of the density of water vapor at STP which would be roughly 1/60 x 1.2 x 18/29 = 12 ug/ cm$^3$.
The mass flow due to diffusion would be $12 \cdot \pi \cdot 5^2 / 5 \approx 0.2 mg / s$. That is surprisingly close to the value we were getting - so we can assume that we don't need air flow to sustain the evaporation. Note that the assumptions about dimensions were pretty much pulled out of thin air...
If the above is valid, then we rule out the need for air flow in the vicinity of the cup (although it could be there, the assumption of "no air flow" will give the largest temperature difference. And that means we need to compute the rate at which heat flows into the water from air.
There are three distinct surfaces: the surface that the cup stands on, the wall of the cup, and the surface of the liquid. It is hard to find a good value for the thermal conductivity of a "cup", so we will use $20 W/mK$ as a representative number for a not-very-conductive ceramic. For the same cup as above, with water to a depth of 5 cm, the surface area is approximately 150 cm$^2$, and for a thinness of 4 mm we get a thermal resistance of 75 W/K. Now we need to add the thermal resistance of the air around the cup: the thermal conductivity of air is quite small at 0.024 W/mK so it will dominate the calculation. At the surface of the cup we would need a thermal gradient of 0.2 / (0.015\cdot 0.024) = 550 K/m which would be very significant - and in fact it would create a sufficiently large density gradient that convection would be set up.
If we assume a low air velocity, we find (from engineeringtoolbox.com) the h factor around 15 W/m$^2$K; for the same parameters as before, this gives a thermal conductivity of about 0.2 W/C, which implies a temperature difference of about 1C is needed to provide the necessary heat flow (0.2 W, as you had calculated). This is much less than the conductivity of the ceramic, which can therefore safely be ignored. We will also ignore thermal conduction to the liquid surface because the air above the liquid was assumed to be stagnant (talk narrow cup assumption). It's much easier to set up convection on the periphery of the cup rather than across the top.
Note that this calculation is quite sensitive to some very approximate assumptions, so the result can be wildly off. But is demonstrates the principle - the liquid would be about one degree C cooler than the surroundings if the air is quite steady at 20C and 50% relative humidity.