If we aren't concerned with exactitude and are willing to accept some $\pm\epsilon$ then it is merely a question of "how much" error we can tolerate.
One can mark down halves successively to get lengths of $\frac{1}{2^n}$
Once you get that measurement as an acceptable unit, you can count $x$ units off to get to $\frac{1}{3}$ and $\frac{2}{3}$
Let me walk you through a simple example:
Say the length of the block is 100.
Then we want to cut it off at 33.33 and 66.67.
Now successively measure halves (half of half of half of...) 4 times. This gives us a unit of $\frac{1}{16}*100 = 6.25$
Now we can count off 5 units as 31.25, and 6 units as 75, which are "pretty close" to 33.33 and 66.67 respectively.
If your error tolerance is low, you can just keep making the unit smaller till you fit inside your $\pm\epsilon$ range.
In the above example, if you had successively measured halves 8 times, instead of 4, the unit would have been $\frac{1}{256}*100 = 0.390625$
Counting off 85 units = 33.20 (yup, really close to 33.33)
Counting off 171 units = 66.79 (reasonably close to 67.67)
In fact you can do one better. After counting off 85 units and cutting that piece off, you are left with 66.80. Now just chop that off into half, and you'll end up with 3 pieces: 33.20, 33.40, 33.40
The benefit of using the length as 100 is this: all the numbers above can actually be expressed as %.
Now if we had to issue a blanket statement, as to how many successive measures of half it takes to arrive at the right unit, you could say
$max(\frac{1}{2^n} l) < 2 \epsilon $
expanded from David's comment
$f' > f$ means $f'/f > 1$ so $(\log f)' > 1$. Why not take $\log f > x$, say $\log f = 2x$, or $f = e^{2x}$.
Thus $f' > f > 0$ since $2e^{2x} > e^{2x} > 0$.
added: Is there a sub-exponential solution?
From $(\log f)'>1$ we get
$$
\log(f(x))-\log(f(0)) > \int_0^x\;1\;dt = x
$$
so
$$
\frac{f(x)}{f(0)} > e^x
$$
and thus
$$
f(x) > C e^x
$$
for some constant $C$ ... it is not sub-exponential.
Best Answer
If you square $\displaystyle 2$, $\displaystyle k$ times, you get $\displaystyle 2^{2^k}$.
Thus you need
$$ 2^{2^k} \ge n$$
i.e.
$$ k \ge \log_2 (\log_2 n)$$
and so you can pick
$$ k = \lceil \log_2 (\log_2 n) \rceil$$
Where $\displaystyle \log_2$ is log to base $\displaystyle 2$.