This is actually hinted at in the typical Trenberth energy budget diagram for the Earth:
The $86.4\mathrm{\frac{W}{\,m^2}}$ is that energy movement up into the atmosphere from vaporization. The $340.3\mathrm{\frac{W}{\,m^2}}$ is the overall greenhouse effect.
The very heavy majority of that natural greenhouse effect is water vapor. The water cycle indicates that on average that water vapor stays in the atmosphere for two weeks (this article now suggests 8-9 days). So from those details, that means we can consider a water vapor's greenhouse return to be about four times its phase change energy moved. And so a quarter of the residency time works out to a rough estimate that it takes about 2-4 days for the average water molecule to send as much energy back to the Earth as it brought up from surface to the atmosphere by its absorption and eventual condensation/deposition.
I can imagine your friend may not feel comfortable with just using this figure to consider the estimate, so perhaps lets try an alternate way of tunneling down to where these numbers come from....
For steady water vapor levels in the atmosphere, the only major way in the end that moisture transfers energy from surface to atmosphere is by its resultant precipitation (other water transitions tend to have no net change on moisture, and don't shift net energy from the ground... water that evaporates and then recondenses at the ground has no net energy change, likewise water vapor that becomes a cloud, then reevaporates at that level has no net energy change over that process).
The average daily precipitation on Earth is around 2 mm per area. The energy for evaporation is 2256 kJ/kg water. So for a square meter, that's
$$0.2 \frac{\,\mathrm{cm}}{\,\mathrm{day}} \cdot 1\,\mathrm{m^2} \cdot \left(\frac{1\,\mathrm{g}}{\,\mathrm{cm^3}}\right) \cdot 2256 \frac{\mathrm{J}}{\mathrm{g}} \cdot \left(\frac{100\,\mathrm{cm}}{1\,\mathrm{m}}\right)^2 = 45120 \frac{\mathrm{J}}{\mathrm{m^2}}\;\mathrm{per}\;\mathrm{day}$$
(the third term is the density of water of $\frac{1\,\mathrm{g}}{\,\mathrm{cm^3}}$ and the fifth term matches units)
Which, converting to per second to get in Watts (multiply by $\mathrm{\frac{1\,day}{86400\,sec}}$), that's about $.5\mathrm{\frac{W}{\,m^2}}$, which comes out to be on roughly the same scale of value as the figure's latent heat value.
How long does it take to achieve that in greenhouse reradiation by that particular water vapor? In a sense it's not easy to calculate the energy directly by each molecule... we know the radiative properties of water, what wavelengths water reflects/absorbs/radiates more at, but it's a cumulative distribution, each molecule affected by its location and interaction with the others that determines the overall amount. We can look how the greenhouse effect roughly doubles the energy we receive in a day, meaning half of all surface energy is from previous surface radiation being returned back... which works out to be that $340.3\mathrm{\frac{W}{\,m^2}}$ of greenhouse return). But either way, it all leads back to residence time meaning it's that same ratio that was given from the figure, 86.4/340.3, and so the same time. It may really help to think of it not so much as how long it stays and how much that water molecule contributes while it is in the atmosphere, but instead about the whole water cycle process being what maintains the amount of water vapor in the atmosphere to allow for that amount of continued greenhouse return, and so it's just one continuous amount versus the other. But if you really do want to think of it as the time for water vapor to generate its greenhouse effect matching its latent heat transfer, it looks like it must be on the order of about 3 days, or around a quarter of its time in the atmosphere, to result in the energy amounts we see.
