This is an exercise from Abbott's real analysis book. It's exercise 3.4.4.(b) on page 93. I couldn't find a definition of ''dimension'' in the book. The only thing I could find is something on page 77. It seems to say that if you "magnify" an object by a factor of $3$ and denote the dimension by $d$ then $3^d = $ size of the magnified object. Example: if the object is a square of length $1$ then scaling it by $3$ is $9$ times the square $\implies$ $3^d = 9$ $\implies d=2$. I tried to apply this to this exercise:
Let $C$ now be the fat Cantor set: Let $C_0 = [0,1]$, $C_1= [0,3/8] \cup [5/8,1]$ and so on (remove the middle fourth in each step).
$C = \bigcap_n C_n$ is closed since it is an intersection of closed sets and it is bounded since it is contained in $[0,1]$ therefore $C$ is compact. It is perfect because it is closed and because it does not contain any isolated points: if $x \in C$ then for every $n$, $x \in C_n$ and since the endpoints of $C_n$ are in $C_n$ there is a point other than $x$ in $B(x, 2/8(2/3)^n) \cap C$.
The length of $C$ is $1$ minus everything that is removed: $$ 1- 1/4 - 2/8 \sum_{k=0}^\infty (2/3)^k = 1-1/2 = 1/2$$
Its dimension is computed by starting with $[0,3]$ and noting that removing the middle fourth yields two intervals of length $11/8$. Using these to produce two Cantor sets yields two Cantor sets of length $11/8$ each. Therefore the dimension of the new Cantor set (Cantor set produced from $[0,3]$) is
$$ 3^d = 2 \times {11 \over 8} = {22 \over 8}$$
solving for the dimension of the Cantor set $d$:
$$ d = {\log {22\over 8}\over \log 3} = {\log 22 - \log 8 \over \log 3}$$
and I don't know how to simplify this expression further. Is this $d$ the dimension of the fat Cantor set?
Edit I founbd this. According to this the Hausdorff dimension should be $1$. So my calculation should be false.
I also found this. According to this, if $N$ is the number of self similar copies and $s$ is the scaling factor then $N= s^d$. I computed that there are ${22\over 8}$ self similar copies. So my calculation should be correct.