Resource allocation in self-replicative systems

The total mass of the system is $\mathbf{S} = \mathbf{S}_O + \mathbf{S}_R$

The system is in exponential growth, so we have $\dot{\mathbf{S}} = \lambda \mathbf{S}$

We also know that the replicative part of $\mathbf{S}$ is $\mathbf{S}_R$ , and each unit mass of $\mathbf{S}_R$ can produce $\gamma$ units of $\mathbf{S}$ per unit time.

We therefore also have $\dot{\mathbf{S}} = \gamma \mathbf{S}_R$

Equating the two expressions, we find

$\begin{aligned} \lambda \mathbf{S} &= \gamma \mathbf{S}_R \\ \lambda / \gamma &= \frac{\mathbf{S}_R}{\mathbf{S}} \\ \lambda / \gamma &= \frac{\mathbf{S}_R}{\mathbf{S}_R + \mathbf{S}_O} \end{aligned}$

We're told that the system doubles in size once every 2 hours. The solution to $\dot{\mathbf{S}} = \lambda \mathbf{S}$ is given by $\mathbf{S}(t) = \mathbf{S}_0 e^{\lambda t}$ Therefore, $\displaystyle\lambda = \frac{\log 2}{2} \mbox{ hour}^{-1}$

We then have the mass fraction of $\mathbf{S}_R$ in the system, $f_R$ given by $\displaystyle \frac{\mathbf{S}_R}{\mathbf{S}} =\lambda/\gamma = \frac{\log 2}{6} \frac{\mbox{kg }\mathbf{S}_R}{ \mbox{kg } \mathbf{S}} \approx 11.55\%$

Using the constraint $\mathbf{S} = \mathbf{S}_O + \mathbf{S}_R$ we find $f_O = 1-f_R$ and thus $\boxed{f_O = 88.45\%}$ .