Let's consider a sphere of radius $R$ at $T_0=100$ C in a far field of water with temperature $T_\infty=30$ C Let's consider a sphere of radius $R$ at $T_0=100$ C in a far field of water with temperature $T_\infty=30$ C.
How to calculate the sphere's temperature in each "layer" at any time?
I really don't know how to solve this problem,
I know the heat equation:
$$Q=mcT$$
$m$: is the object mass
$c$: Is the object specific heat
$T$: Is the object temperature change
and the Newton's Law of Cooling
$$\frac{dT}{dt}=-k(T-T_a)$$
where $k=\dfrac{\alpha S}{mc}$
$S$: is the object surface area
$\alpha$: Is the a constant
 A: Newton's Law of Cooling actually states:
$$\frac{dQ}{dt}=-kA[T(t)-T_a],$$
where $\frac{dQ}{dt}$ is the heat flux, $k$ the heat transfer coefficient, $A$ the sphere's surface area, $T(t)$ the sphere's temperature as a function of time and $T_a$ the ambient temperature (assumed constant).
When the sphere drops in temperature by $dT(t)$ then it loses an amount of heat $dQ$:
$$dQ=mcdT(t),$$
where $m$ is the mass of the sphere and $c$ the heat capacity of the sphere's material (assumed constant).
Combining we get:
$$\frac{mcdT(t)}{dt}=-kA[T(t)-T_a]$$
Reworked:
$$\frac{dT(t)}{T(t)-T_a}=-\alpha dt$$
With:
$$\alpha=\frac{kA}{mc}$$
Integrated between $t=0$, $T=T_0$ and $t$, $T$, we get:
$$\ln \frac{T-T_a}{T_0-T_a}=-\alpha t$$
So:
$$\large{T=T_a+(T_a-T_0)e^{-\alpha t}}$$
Note that only for $t \to \infty$ does $T \to T_a$.
Inversely we can write:
$$\large{t=\frac{1}{\alpha}\ln \frac{T_0-T_a}{T-T_a}=\frac{mc}{kA}\ln \frac{T_0-T_a}{T-T_a}}$$
This is the predicted temperature evolution, assuming the sphere's temperature is homogeneous at all times. Values for $k$ can be obtained from various thermal engineering websites.
Specifically for large spheres that is not true, as the core temperature will remain higher than the surface temperature and there's a radial temperature gradient $\frac{dT}{dR}$.
So with regard to:

How to calculate the sphere's temperature in each "layer"? at any time

This derivation is analytically much harder and can be found here.
Basically a system of two simultaneous differential equations has to be solved:
$$\frac{\partial T}{\partial t}=a \frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial T}{\partial r})$$
$$-b\frac{\partial T}{\partial t}=h(T-T_a)$$
with $a$ the thermal diffusivity, $b$ the thermal conductivity and $h$ the heat transfer coefficient.
A: Let's assume that the initial temperature of the sphere is not 1000C, so that, at least, it is not above the critical temperature of water.  Let the initial temperature be $T_0$ and let $T_{\infty}$.  Let's also assume that the pressure is high enough so that the water close to the sphere does not boil.
My goal will be to arrive at an upper bound to the time for the average sphere temperature to drop to a given value.  So I will be neglecting natural convection heat transfer, and focusing entirely on stagnant conductive heat transfer.
The heat transfer coefficient at the surface of the sphere on the water side of the sphere surface will start off very high, and then gradually decrease to an asymptotic value at long times, given by $k_w/R$, where $k_w$ is the thermal conductivity of water and R is the outer radius of the sphere.  In a similar way, the heat transfer coefficient at the surface on the sphere material side of the boundary will also approach an asymptotic value at long times, given by $2k_S/R$, where $k_S$ is the thermal conductivity of the sphere material.
Based on these asymptotic values, the overall heat transfer coefficient between the water (far field) and the sphere will be given by:$$\frac{1}{U}=\frac{R}{k_w}+\frac{R}{2k_S}$$ 
Solving for U gives:$$U=\frac{2k_Sk_w}{R(2k_S+k_w)}$$
From a heat balance on the sphere, we have:
$$(\frac{4}{3}\pi R^3)\rho C_p\frac{dT}{dt}=-4\pi R^2U(T-T_{\infty})$$
Substituting our equation for the overall asymptotic heat transfer coefficient into this equation yields:$$\frac{dT}{dt}=-\frac{(T-T_{\infty})}{\tau}$$where $\tau$ is the characteristic cooling time, given by:
$$\tau=\frac{\rho C_p R^2(2k_S+k_w)}{6k_Sk_w}\tag{1}$$
Eqn. 1 for $\tau$ provides a value of the parameter $\alpha$ (or k) that was alluded to but not provided in the previous answers.
