i got the number of iteration,$n$, to achieve the accuracy, $\epsilon=10^{-2}$ is $n=5.5\approx 6$
But in answer script, $n=8$.
My procedure is
$ \frac{(b-a)}{2^n}<\epsilon$
$\Rightarrow\frac{(3.2-1)}{2^{n}}<10^{-2}$
$\Rightarrow{(2.2)}{2^{-n}}<10^{-2}$
Taking $\log_{10}$
$\log_{10}(2.2)-n\log_{10}(2)<-2$
$\Rightarrow n>\frac{2-n\log_{10}(2)}{\log_{10}(2.2)}=5.5\approx 6$
Where is the mistake?
You want $\frac{2.2}{2^n} < 10^{-2}$, or in other words, $220 < 2^n$. Then $\log_2 220 < n$. Since $\log_2 220 \approx 7.7814\cdots$, we see that the smallest such $n$ is $n=8$.
Alternatively, you could notice that $f(3) = 0$., and so $n=0$ :-).