The issue seems debated, since the two most upvoted answers give opposite conclusions. I decided to carry out the full calculation, hoping to settle the matter.
Let's settle the scene: we built a giant hollow spherical structure, made of a very resistant and rigid material. We place the sphere in empty space and we measure its circumference with a yardstick and find it is equal to a certain value, let's call it $2 \pi \mathcal{R}$. Then we measure the radius and we find it is equal to $\mathcal{R}$. All good, we have proven that empty space is flat. Nice to know. So, in empty space, the metric is given by
$$ds^2 = - dt^2 + dr^2 + r^2 d\theta^2 + r^2 \sin^2(\theta)d\phi^2$$
and the volume of the sphere is
$$V = \int_0^\mathcal{R} dr \cdot r d\theta \cdot r \sin(\theta) d\phi = 4 \pi \int_0^\mathcal{R} r^2 dr = {4 \pi \over 3}\mathcal{R}^3$$
Now we fill the sphere uniformly with gas that has total mass $M$. We need not to lose our sleep in wondering how this gas can be kept at uniform density, despite its self gravitational pull, for this is irrelevant for this question: as you will see in a moment, pressure and temperature of the gas are not involved in the spacial part of the metric.
The general expression of the metric of a spherically symmetric matter distribution can be found in this stack exchange answer
$$ds^2 = - B(r) dt^2 + A(r) dr^2 + r^2 d\theta^2 + r^2 \sin(\theta)d\phi^2$$
where
$$A(r) = \left(1-{2G m(r)\over r}\right)^{-1}$$
and $m(r) = 4\pi \int_0^r r'^2 \rho(r') dr'$. In our case, $\rho$ is constant inside the sphere and zero outside, so that $m(r) = M {r^3 \over \mathcal{R}^3}$ when $r<\mathcal{R}$ and $m(r) = M$ when $r\geq \mathcal{R}$.
Inside the sphere, $A_{in}(r) = \left(1-{2G M \over \mathcal{R}} {r^2 \over \mathcal{R}^2}\right)^{-1}$ and outside the sphere $A_{out}(r) = \left(1-{2G M \over r} \right)^{-1}$, as required by the usual Schwarzschild metric. As already noted by BarrierRemoval, $A>1$, and this means that the volume inside the sphere is increased. But let's proceed slowly.
Now if we measure the circumference again, we will still obtain $2 \pi \mathcal{R}$, because the tangential part of the metric has not changed. Instead, if we went and measured the radius we would obtain
$$\int_0^\mathcal{R}\left(1-{2G M \over \mathcal{R}} {r^2 \over \mathcal{R}^2}\right)^{-1} dr> \mathcal{R}$$
The space is curved! And the volume is
$$V = 4\pi \int_0^\mathcal{R} r^2 \left(\sqrt{1-{2G M \over \mathcal{R}} {r^2 \over \mathcal{R}^2}}\right)^{-1} dr > {4 \pi \over 3} \mathcal{R}^3$$
The volume has increased. Now imagine we compress the gas (but keeping our rigid sphere as it is) until it becomes a small ball at the center of our sphere, with uniform density, mass $M$ and radius $L$. Let's pretend that this homogeneous ball of gas represents a star. We already know how to calculate the volume inside our spherical rigid structure:
$$V = 4\pi \int_0^L r^2 \sqrt{A(r)_{in}} dr + 4\pi \int_L^\mathcal{R} r^2 \sqrt{A(r)_{out}} dr$$
The integral is actually solvable analytically (check it in Mathematica), but I won't write the solution here in matjax, because it is quite long. But I have implemented it in python and I leave you the script
import numpy as np
import matplotlib.pyplot as plt
# rs=2GM is the Schwarzschild radius
rs = 1
# the radius of the rigid sphere
R = 5
# the radius of the gas ball
L = np.linspace(1.001,R,1000)
def V(rs,L,R):
# calculate the volume of the sphere
return 4*np.pi*(I1(L,L)-I1(0,L) + I2(R)-I2(L))
def I1(r,L):
# primitive of the internal integral
return 1/2*(L**(9/2)/rs**(3/2)*np.arcsin(np.sqrt(rs/L**3)*r) - L**3*r/rs * np.sqrt(1-rs*r**2/L**3))
def I2(r):
# primitive of the external integral
return 2*rs**3*np.sqrt(1-rs/r)* (5*np.arctanh(np.sqrt(1-rs/r))/(16*np.sqrt(1-rs/r)) - r**3/(48*rs**3) * (-15*(1-rs/r)**2 + 40 *(1-rs/r) - 33))
plt.plot(L,V2(rs, L, R)/(4*np.pi*R**3/3))
plt.xlabel("$L$")
plt.ylabel("$V/V_0$")
plt.grid()
plt.show()
The figure here shows how ${V \over 4\pi/3\mathcal{R}^3}$ varies as a function of $L$, where the other parameters have been fixed at $r_s=1$ and $\mathcal{R} = 5$. If $L \to r_s$ the uniform gaseous sphere gets close to become a black hole and the volume increases. But I don't think it diverges, I believe it tends to a finite value, although I have not attempted to prove it.
