 Solution

Let the surface be at $R$ with vanishing external pressure. Multiply hydrostatic equilibrium by $4\pi r^3$ and integrate:The other side is $-{\int }_0^{R}4\pi Gm(r)\rho (r)rdr=-{\int }_0^{M}Gm/r(m)dm$. Building the star one shell at a time identifies this as its negative gravitational potential energy $E_{\mathrm{grav}}$. ThereforeIf the star has nonzero surface pressure $P_s$, the integration instead gives $3⟨P⟩V-3P_{s}V=-E_{\mathrm{grav}}$; the requested form assumes $P_s=0$.