Initially let pressure is $p$, then

$p V=n_1 R T$

Finally pressure is $p$ and volume is $\frac{V}{3}$.

Let number of moles of gas left is $n_2$, then

$p \frac{V}{3}=n_2 R T$

Dividing both equations, we get

$\frac{n_2}{n_1}=\frac{1}{3} \text { or } 1-\frac{n_2}{n_1}=1-\frac{1}{3} \text { or } n_1-n_2=\frac{2}{3}$

Hence, $\frac{n_1-n_2}{n_1} \times 100=\frac{2}{3} \times 100=66 \%$ or percentage of number of molecules escaped $=66 \%$.