Please consider me a beginner and not an expert by any means. My question below will thus be very very simple as that of an uninitiated. Let me consider four simple physical systems with ${\rm N}_S$ states. With those, I'll try to explain my current state of knowledge/understanding at this point and ask the question.
First, one coin. It has two possible states- a 'head' and a 'tail' i.e., ${\rm N}_S=2$. Each state can be denoted by $0$ and $1$. This is an example of a $1$-bit system.
Second, consider $N$ identical coins. Clearly, the number of states (or configurations) is now $N_S=2^N$. Each state of the full system can again be represented or encoded by a distinct string of '$p$' zeros and '$q$' ones such that $p+q={\rm N}$. This is an example of a classical ${\rm N}$-bit system.
Third, consider a dice where the number of states $N_S=6$. By definition, this is a $\log_2 6\approx 2.585$-bit system.
Fourth, consider $N$-dies so that the number of states is $N_S=6^N$. It is, by definition, a $\log_2(6^N)\approx 2.585N$-bit system.
Therefore, irrespective of whether each 'microscopic' constituent is a $1$-bit system (e.g., a coin) or not (e.g., a dice), the quantity $X=\log_2N_S$ is used define a $X$-bit system. I have tried to explain that with my four examples above.
Given the above set up, my question is, if we have a $X$-bit system, what is the amount of information carried by that system?