How to calculate the radius of curvature of a FLRW universe in terms of energy density $\rho$? It is easy to calculate the radius of curvature $R_K$ using critical density ${\rho}_{\text{cr}}$ which depends on the Hubble parameter $H$. However, I would like to know how I can calculate the radius of curvature at a given time $t$ in terms of the energy density budget of the Universe, i.e. radiation density ${\rho}_{R}$, matter density ${\rho}_{M}$, and dark energy density ${\rho}_{\Lambda}$. 
I am interested in calculating the curvature of an FLRW-universe, not necessarily our own Universe. 
 A: The radius of curvature in an FLRW-universe is
$$R_k=\frac{c}{H \sqrt{-Ω_k}}$$
so you need the Hubble parameter
$$H=H_0 \sqrt{\Omega_R \ a^{-4}+\Omega_M \ a^{-3}+\Omega_K \ a^{-2}+\Omega_{\Lambda}}$$
and the curvature parameter
$$\Omega_k= \frac{\rho_k}{\rho_c} \ , \ \  \rho_k = \frac{\rho_K}{a^2} \ , \ \  \rho_K=\Omega_K \ \rho_c \  , \ \ \rho_c=\frac{3H^2}{8 \pi G}$$
The capitalized subscripts represent the present values, and the lowercase ones the scale-factor-dependend values. The present curvature parameter is $\Omega_K=1-\Omega_T$ where
$$\Omega_T=\Omega_R+\Omega_M+\Omega_{\Lambda}$$
The time by scale factor is
$$t=\int_0^a \frac{1}{a' \ H} \, {\rm d}a'$$
and in general the scale factor by time has to be solved numerically, unless you assume flatness and neglect the radiation, then the inverse function can be solved analytically.
So to find your curvature radius for a given time $t$ first find the corresponding scale factor $a$, plug it into the equations for $H$ and $\Omega_k$ and calculate $R_k$.
The distance from yourself to yourself is the circumference $C=2 \pi R_k$ and the proper volume of the closed space is the area of the corresponding hypersphere $V=2 \pi^2 r^3$
