= Solution
For conserved nonrelativistic matter, $\rho_M=\rho_{M0}a^{-3}$, where $a=R/R_0$ and $a_0=1$. Substitution into the <Friedmann equation> gives
$$
H^2=H_0^2\left(\Omega_Ma^{-3}+\Omega_ka^{-2}+\Omega_\Lambda\right),
$$
with
$$
\boxed{\Omega_M=\frac{8\pi G\rho_{M0}}{3H_0^2},\qquad
\Omega_k=-\frac{kc^2}{H_0^2R_0^2},\qquad
\Omega_\Lambda=\frac{\Lambda}{3H_0^2}.}
$$
The source writes the dimensionless normalized factor as $R$ in its final expression; this is equivalent to setting $R_0=1$ consistently in the curvature coefficient. Evaluating today gives $\Omega_M+\Omega_k+\Omega_\Lambda=1$. Here the printed $\Lambda$ already has dimensions of inverse <time> squared. If the <cosmological constant> is instead defined geometrically with inverse-length-squared units, replace this $\Lambda$ by $\Lambda_{\rm geom}c^2$.
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