If these are perfectly independent systems, then the total number of eigenstates is a simple product of $H_1$ and $H_2$. You can imagine this as a giant square where one axis as $H_1$ positions and the other has $H_2$ positions. So the total number of positions is:

$$10^{20} \times 10^{22} = 10^{42}$$

That is under the assumption that objects can not be superposed (e.g. we are ignoring the possibility of combinations of states in $H_1$ and $H_2$)

The entropy is then just the logarithm of the number the states times a proportionality constant:

$$S=k \ln \Omega$$

If we set $k=1$ then

$$S=\ln 10^{42} = 96.7$$

We could also use $log_{10}$, in which case the entropy is just $42$.

If these are perfectly independent systems, then the total number of eigenstates is a simple product of $H_1$ and $H_2$. You can imagine this as a giant square where one axis as $H_1$ positions and the other has $H_2$ positions. So the total number of positions is:

$$10^{20} \times 10^{22} = 10^{42}$$

That is under the assumption that objects can not be superposed (e.g. we are ignoring the possibility of combinations of states in $H_1$ and $H_2$)

The entropy is then just the logarithm of the number the states times a proportionality constant:

$$S=k \ln \Omega$$

If we set $k=1$ then

$$S=\ln 10^{42} = 96.7$$

We could also use $log_{10}$, in which case the entropy is just $42$.

I would add that Mark Mitchison's answer is more correct in the way that you would calculate this, but because of the simple case proposed this would work as well.