Solution:

Let $x^2 + y^2 = 64$ be circle $O,$ $x^2 + y^2 = 4$ be circle $P,$ and $(x - 5)^2 + y^2 = 3$ be circle $Q.$

First note that every circle, $R,$ in $S$ is internally tangent to $O.$ Then we case on the tangency of $R$ with $P$ and $Q.$

Case $1:$ This corresponds to the pink circle. This is where $P$ and $Q$ are internally tangent to $R.$

Case $2:$ This corresponds to the bluish circle. This is where $P$ and $Q$ are externally tangent to $R.$

Case $3:$ This corresponds to the green circle. This is where $P$ is externally and $Q$ is internally tangent to $R.$

Case $4:$ This corresponds to the red circle. This is where $P$ is internally and $Q$ is externally tangent to $R.$

We can consider cases $1$ and $4$ together. Note that $O$ and $P$ have the same center. This means that the line connecting the center of $R$ and $O$ passes through the tangency point of both $S$ and $O$ and $S$ and $P.$

This line is the diameter of $R,$ and it has length $$r_P + r_O = 2 + 8 = 10.$$ Therefore, the radius of $R$ is $5.$

Consider cases $2$ and $3$ together. Similarly to above, the line connecting the center of $R$ and $O$ will pass through the tangency points.

This time, however, the diameter of $R$ is $$r_P - r_O = 8 - 2 = 6.$$ This makes the radius of $R$ $3.$

$S$ contains $8$ circles: $4$ of which have radius $5$ and $4$ of which have radius $3$ (this is because we can flip all the circles in the diagram over the x-axis to get $4$ more circles).

The total area of the circles in $S$ is therefore $$4 (5^2 \pi + 3^2 \pi) = 136 \pi.$$ Thus, E is the correct answer.