As every system you can write down an energy balance to compute the amount of heat that has to be given to keep your water at constant temperature.
$$\dot{Q}_{fire} = \dot{Q}_{losses}$$
Now your energy losses are represented by the energy transported by the vapour that flows out of your container and heat exchanges with the environment:
$$\dot{Q}_{losses} = \dot{m}_{vap} l_h + hA(T_w - T_{env})$$
$\dot{m}_{vap}$ being the amout of vapour escaping from the system (kg/s), $l_h$ the water latent heat. For the second term, $h$ (W/m2K) may be some equivalent of an exchange coefficient. The exchange takes place at the surface ($A$) between the water and the surroundings. In the end, $T_w = 373$ K and $T_{env}$ could be the temperature of the surrounding environment. Now $h$ is quite problematic to calculate, because it depends on a lot of factors that would make this explanation quite tedious.You can reasonably assume $h = 50$ W/m2k even though it may vary between 10 and 100.
Edit, in response to your comment,
I wondered if you could provide some more insight on the exchange coefficient as some sources claim that for boiling water the heat transfer coefficient ranges between 3,000-100,000 W/m2K.
Of course I can get in to the details of the exchange coefficient. I would say that the order of magnitude you find out are those typically found for water vapour in relatively "extreme" conditions such as those you can find in electric generation power plants. I was thinking more about the liquid water itself. The exchange coefficient is based on several properties, such as radiation, the fluid motion (convection) and thermal conductivities of everything is in contact with the water (container, air). It also depends on the geometry of the containter. Each of these mechanisms are object of University courses that I can never summarize here in a simple post.
However I think there is a more simple and intuitive way to find out what could be a reasonable $h$.
Imagine that you stop the fire at a given instant $t_0$. From that moment the temperature of the water would change following this simple differential equation:
$$mc_p\frac{\partial T}{\partial t} = hA(T_{env} - T).$$
We neglect the loss of vapour for the moment, since they do not change the reasoning.
The solution we search is in the form of
$$T(t) = T_{env} + (T(t_0) - T_{env})e^{-\frac{hA}{mc_p} t}.$$
At this stage you start to ask yourself how much could it take for the water to cool down to ambient temperature? You can do a guess or perform an experiment. In any case, you put your estimated $t$, togheter with the well known $m$ and $c_p$ and solve for $h$. I let you do the math. Let me know if $h \approx 10$ or $\approx 10^4$!.
