How is a calorie related to the Boltzmann constant? A (small) calorie is defined as the energy needed to increase the temperature of $1g$ of water by $1\space K$, and its value is $1\space cal = 4.184\space J$.
How is this empirical value theoretically related to the Boltzmann constant $k_B$ ?
The Boltzmann constant $k_B=1.380649 \cdot 10^{−23}J.K^{-1}$ now defines the Kelvin unit, relating the temperature (in $K$) with the corresponding average microscopic kinetic energy (in $J$).
Then, knowing the temperature and the number of particles (from the mass and the molar mass) we could calculate the total microscopic kinetic energy : for $1\space g$ of water at $1\space K$, shouldn't this energy value be equal to $1\space cal$ ?
From my attempts, it isn't : the calculated  total kinetic energy of $k_B\cdot (N_A/18.01528) \approx 0.4615\space J$  is about 10 times lower than $1\space cal = 4.184\space J$.
 A: 
From my understanding, the Boltzmann constant $k_B=1.380649 \cdot 10^{−23}J.K^{-1}$ states that $1K$ of temperature is equals to an average kinetic energy of $1.380649 \cdot 10^{−23}J$ on the molecules.

This is not correct. The Boltzmann constant doesn't really "state" anything - it is a conversion factor between absolute temperature and energy.  The most fundamental definition of temperature which arises from statistical mechanics tells us that if we hold the external parameters (e.g. volume, particle number, etc) of an isolated system fixed, then any infinitesimal change in (dimensionless) entropy $\delta \sigma$ is accompanied by a corresponding infinitesimal change in energy $\delta U$ given by
$$\delta U = \tau \delta \sigma\tag{1}$$
for some quantity $\tau$.  The dimensionless entropy $\sigma$ is simply the log of the number of available states in the system, and is a pure number; as a result, its easy to see that $\tau$ has dimensions of energy.
However, thermodynamics was developed as an empirical science well before its fundamental roots in statistical physics were understood.  Historically, it was observed that when you place a hot object into contact with a cold object, then energy would flow from hot to cold until equilibrium was reached. The idea of temperature quantified this - for two objects $A$ and $B$, $T_A>T_B$ meant that $A$ was hotter than $B$.  Scales for temperature were developed based on human body temperature and the freezing point of brine or the on the freezing and boiling point of pure water. Eventually it was understood that the temperatures of most systems had a common minimum possible value, which led to the development of the concept of absolute temperature, whose zero point corresponded to this value.
Once it is understood that thermodynamics can be essentially derived as a limiting case of statistical mechanics, $\tau$ can be identified with the absolute temperature of the system. However, because the absolute temperature which arises in thermodynamics was not defined with dimensions of energy, we instead define a constant $k_B$ with units of energy over absolute temperature which acts as a conversion factor between the fundamental quantity $\tau$ and the empirically understood quantity $T$ via $\tau = k_B T$.  We similarly define $S\equiv k_B \sigma$ to be the (dimensionful) entropy, and so $(1)$ takes the form
$$\delta U = T \delta S \tag{2}$$
which is the definition of entropy which initially arose from thermodynamics.

Now it turns out that for a non-relativistic, monatomic ideal gas in 3 dimensions - for which $U$ is entirely composed of kinetic energy - one finds that
$$U = \frac{3}{2}N \tau \implies \overline u \equiv \frac{U}{N} = \frac{3}{2}\tau \tag{3}$$
Framing this in terms of the familiar $T$, we find that $\overline u = \frac{3}{2} k_B T$ and that a temperature change of $1\ $K corresponds to an energy change of $\frac{3}{2} k_B$ (with $k_B$ measured in energy per kelvin). One could therefore say that the Boltzmann constant is numerically equal to $2/3$ of the amount of energy per atom required to raise the temperature of a non-relativistic, monatomic ideal gas by $1$ K.  The point is that this is a result which arises in an incredibly simple special case, and is far from the definition of the Boltzmann constant.

A (small) calorie is defined as the energy needed to increase the temperature of $1g$ of water by $1K$, and its value is $1\space cal = 4.184\space J$.
I'm unable to calculate this value from the Boltzmann constant $k_B$.

What you're essentially asking for is the specific heat capacity of liquid water. Already this is ill-defined, because that depends on whether the volume or pressure is held constant, and is also temperature dependent:

source
Even if you specify that e.g. you're looking for the isobaric specific heat capacity at $T=10^\circ$ C, the calculation is highly non-trivial.  Unlike an ideal gas, whose particles have no interactions whatsoever, the particles in a liquid are strongly interacting and possess both kinetic and potential energy. Water in particular is even more difficult to handle than usual because of the strong dipolar interactions which result from its large permanent electric dipole moment (i.e. hydrogen bonds).
If it's currently possible to get a good theoretical estimate of the specific heat capacity of water (based on the details of the H$_2$O molecule) then I don't know how to do it. This is far from my field, so I wouldn't be the person to ask. But it's certainly not the case that it could be trivially read off from a proportionality constant.
A: 
As the molar mass of water is $18.01528 g.mol^{-1}$, then $1g$
of water is composed of $(N_{A}\cdot18.01528)$ molecules.

I disagree with this calculation.
With

*

*mass: $m=1\text{ g}$

*molar mass: $m_\text{mol} = 18.01528\ \text{g}\cdot\text{mol}^{-1}$

*Avogadro's constant: $N_A=6.022\cdot 10^{23}\text{ mol}^{-1}$
I get the number of molecules as
$$N = \frac{N_A m}{m_\text{mol}}
= \frac{6.022\cdot 10^{23}\text{mol}^{-1}\cdot 1\text{ g}}{18.01528\ \text{g}\cdot\text{mol}^{-1}}
= 3.34\cdot 10^{22}$$
which is actually a pure number (i.e. without any unit) as it should be.
My resulting number is smaller than yours by a factor of $18^2$.

Using this number $N$ you get
$$Nk_B\ \Delta T
= 3.34\cdot 10^{22} \cdot 1.38\cdot 10^{-23}\text{J}\cdot\text{K}^{-1}\cdot 1\text{ K}
= 0.462\text{ J}$$
The formula for the kinetic energy
$$\Delta E_\text{kin}= \frac{3}{2}Nk_B\ \Delta T$$
is only valid for mono-atomic gases (e.g. helium).
For other ideal gases it is
$$\Delta E_\text{kin} = \frac{f}{2}Nk_B\ \Delta T$$
where $f$ is the total number of degrees of freedom
for a single molecule (i.e. translational, vibrational and rotational motion).
For a water molecule (H$_2$O, 3 atoms) there are $f=9$ degrees of freedom.
So for water steam (i.e. gaseous water) we have
$$\Delta E_\text{kin} = \frac{f}{2}Nk_B\ \Delta T
=\frac{9}{2}\cdot 0.462\text{ J} = 2.08\text{ J}$$
But your question is about liquid water, not gaseous water steam.
For a liquid the physics is much more complicated than in a gas.
The liquid molecules are colliding and touching each other all the time.
Hence the approximation of nearly free molecules, only
colliding once in a while, is not applicable anymore.
There is not only kinetic energy, but also potential energy (during collisions).
So we would expect an energy larger than $2.08\text{ J}$.
But we don't know how much larger, because we don't know how much potential
energy to add.
A: Directly not related at all.
Boltzmann constant relates average kinetic energy of gas particles with temperature of gas.
While small calorie in general is related to water specific heat capacity which is $4.2~J~g^{-1}K^{-1}$ (energy needed to raise temperature of 1 gram of water by 1 K)
BTW, there is a couple of alternative small calorie definitions, for example "20 °C calorie" definition sound like

The amount of energy required to warm one gram of air-free water from 19.5 to 20.5 °C at standard atmospheric pressure

which is $$ cal_{20} \approx 4.182~J $$
