= Solution
For a freely jointed <Gaussian chain>, independent isotropic segment vectors have zero mean and coordinate variance $\ell^2/3$. The <central limit theorem> gives the end-to-end vector density
$$
p(\mathbf R)=\left(\frac{3}{2\pi N\ell^2}\right)^{3/2}\exp\left[-\frac{3R^2}{2N\ell^2}\right].
$$
The number of conformations in a fixed small vector-volume element is proportional to this density. Taking its logarithm gives the configurational <entropy>
$$
S(\mathbf R)=S_0-\frac{3k_BR^2}{2N\ell^2}.
$$
This is the vector-constrained entropy appropriate to an anchored strand. A density for the scalar length alone also has a $4\pi R^2$ shell factor and is a different constraint. The approximation requires many segments and extensions well below $N\ell$.
If the internal energy has negligible extension dependence, the <Helmholtz free energy> is $F=F_0+3k_BTR^2/(2N\ell^2)$. Differentiation gives an <entropic force> with effective spring constant $3k_BT/(N\ell^2)$. In an affine isotropic network, averaging over initially oriented strands gives <rubber elasticity>
$$
G=\nu_s k_BT,\qquad E=3G=3\nu_s k_BT
$$
for an incompressible material; $\nu_s$ is the number of elastically active strands per volume. For example a volume-preserving stretch with principal factors $\lambda,\lambda^{-1/2},\lambda^{-1/2}$ has strain-energy density $W=G(\lambda^2+2/\lambda-3)/2$. Its nominal tensile <stress> is $G(\lambda-\lambda^{-2})=3G(\lambda-1)+O((\lambda-1)^2)$, proving the stated small-strain <Young's modulus>.
Chemically, increasing the effective <cross-link> density, for example by changing vulcanization, generally increases the modulus by shortening strands between cross-links. Dangling ends, loops and changes of network structure mean the chemical junction count need not equal $\nu_s$.
The reference area is $A_0=10^{-5}\ \mathrm{m^2}$. Neglect self-weight, thermal expansion and energetic elasticity, keep the same permanent network, and heat quasistatically in the rubbery regime. Then $G$ increases from $20$ to $22\ \mathrm{MPa}$. With a constant suspended load, the exact affine-network balance is
$$
\frac{mg}{A_0}=G(T)(\lambda-\lambda^{-2}).
$$
It gives $\lambda_{300}=1.01662$ and $\lambda_{330}=1.01509$. Thus
$$
\boxed{\text{the lower end rises by approximately }1.53\ \mathrm{mm}.}
$$
Linear elasticity instead gives extensions $16.35$ and $14.86\ \mathrm{mm}$, a $1.49\ \mathrm{mm}$ rise, consistent at this small strain. Incompressibility makes the area $A_0/\lambda$, already included by using nominal <stress>. Heating increases entropic stiffness, so the loaded rubber contracts and performs work on the mass. Strong thermal expansion, changes in cross-linking or leaving the rubbery regime would require additional material data.
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