I want to extend knzhou's comment. In the quantum theory of the electromagnetic (EM) field the number of "photons" is the number of excitations of the EM field with a given wavevector $\mathbf k$ and a given polarization $\lambda$. In classical EM theory, a wave with definite $\mathbf k$ and $\lambda$ is a plane wave, which has infinite extent to begin with (so it would influence all the water molecules). Of course we may build a localized wave, but in order to do this we'll have to add many plane waves and therefore $\mathbf k$ will assume a range of values. So a standing wave does not have a definite $\mathbf k$ since it's a sum of two plane waves travelling in different directions! We already see that the notion of "extent" of a wave with definite values of $\mathbf k, \lambda$ is ill-defined.
Things carry out similarly in the quantum theory of the EM field, but some subtleties kick in. The first thing that changes is how we describe the field: we use an occupation number ket $|n_{\mathbf k, \lambda}\rangle$ that tells us how many photons (excitations) there are with a given $\mathbf k$ and $\lambda$ and the fields become field-operators that act on the ket. You may ask: "if the fields are operators now, how do we associate them with the classical fields (which are numbers)?". Well, we take the expectation value
$$
\langle n_{\mathbf k, \lambda}| \mathbf E |n_{\mathbf k, \lambda}\rangle
$$
which, if you carry out the calculations, will be zero for a state with a definite number of photons (but $\langle \mathbf E^2 \rangle$ will be infinite!). It'll be non-zero if our field-state is a superposition of different occupation numbers, for instance:
$$
|\psi\rangle = c_0|0_{\mathbf k, \lambda}\rangle + c_1|1_{\mathbf k, \lambda}\rangle.
$$
Indeed, the state that resembles a classical coherent wave is called a coherent state, which is a superposition of infinite photon number states. So even our notion of electric field depends on having an uncertainty in the number of photons.
So, in your example which water molecule would a single photon influence? The answer is still all of them, but only one will be excited by the photon! All molecules will feel the EM field of a single photon, but by conservation of energy only one can absorb the photon and excite. It's interesting to note that even in the absence of photons the field can influence the molecules and de-excite them, making them emit more photons.
In conclusion, both classical and quantum treatments of the EM field will not have a well-defined "length" of a wave/photon with definite $\mathbf k, \lambda$, and if you really want a "reasonable" length you'll have to give up the definiteness of $\mathbf k$, and hence have more than one photon.
