= Solution
For $N=300$, the critical <polymer> <volume fraction> is
$$
\phi_c=\frac1{1+\sqrt{300}}\simeq0.05458.
$$
If $v_0$ is the volume assigned to one <amino acid> in the lattice model, the critical <number densities> are exactly
$$
\boxed{n_{m,c}=\frac{\phi_c}{v_0},\qquad
n_{p,c}=\frac{\phi_c}{Nv_0}=\frac{n_{m,c}}{300}.}
$$
The extra physical approximation needed to use the given <radius of gyration> is that a chain in a <poor solvent> is a compact <polymer globule>. Its size then scales as $R_g\sim bN^{1/3}$ with $v_0\sim b^3$, so $Nv_0\sim R_g^3$. This gives the standard scaling estimates
$$
n_{m,c}\sim\frac{N\phi_c}{R_g^3},\qquad
n_{p,c}\sim\frac{\phi_c}{R_g^3}.
$$
With $R_g=3\,\mathrm{nm}$, numerical substitution gives
$$
\boxed{n_{m,c}\sim6\times10^{26}\,\mathrm{m^{-3}},\qquad
n_{p,c}\sim2\times10^{24}\,\mathrm{m^{-3}}.}
$$
Using only the leading large-$N$ value $\phi_c\sim N^{-1/2}$ gives $6.4\times10^{26}$ and $2.1\times10^{24}\,\mathrm{m^{-3}}$, respectively. These are scaling estimates, not geometrically exact densities.
For comparison, a specifically \i[uniform spherical] compact protein has the <radius of gyration of a uniform sphere>, $R_g^2=3R^2/5$, so its physical volume is $V_p=(4\pi/3)(5/3)^{3/2}R_g^3$. Taking $Nv_0=V_p$ instead gives $n_{m,c}\simeq6.7\times10^{25}\,\mathrm{m^{-3}}$ and $n_{p,c}\simeq2.2\times10^{23}\,\mathrm{m^{-3}}$. The factor between these conventions is geometric. The PDF does not give a <monomer> volume, globule density or shape model, and a <radius of gyration> alone does not determine an exact molecular volume. The model-independent conclusions are the formulas in terms of $v_0$ and the 300:1 ratio of <monomer> to protein number density.
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