I have a problem in realizing of this question. As it is mentioned in the picture, we have a cake which is sliced roughly in half, the largest part being chosen each time and the other pieces discarded. I cannot understands the proportions it considers for a random cut:
And why it just considers 2/3 and 3/5 in computing the E[log C1] ? Where are 1/3 and 2/5?
Here's basically what's going on in this problem: At each turn, the cake is replaced by some portion of it, which is either $2/3$ (with probability $3/4$), or $3/5$ (with probability $1/4$). The other pieces are always discarded.
Suppose that it was always $2/3$ (with probability $1$). Then each turn, the cake would be $2/3$ of its previous size, which means that after $k$ turns, it would be $(2/3)^k$ of its original size:
$$ \text{Size}_k = \text{Size}_0 \left(\frac23\right)^k $$
or equivalently,
$$ \frac{\text{Size}_k}{\text{Size}_0} = \left(\frac23\right)^k $$
However, the actual situation is that $1/4$ of the time, the ratio is $3/5$. That means that of the $k$ fractional factors on the right-hand side, about a fourth of them are $3/5$ and not $2/3$. Heuristically,
$$ \frac{\text{Size}_k}{\text{Size}_0} \sim \left(\frac23\right)^{3k/4}\left(\frac35\right)^{k/4} $$
where I intend the $\sim$ to mean that "varies approximately as". If you take the log of both sides, you get
$$ \log \frac{\text{Size}_k}{\text{Size}_0} \sim \frac{3k}{4} \log \frac23 + \frac{k}{4} \log \frac35 $$
and then dividing both sides by $k$ gives you the provided solution (provided you replace the informal $\sim$ with a more rigorous limit demonstration).