Let $n=$ number of moles of gas in both of the bulbs and capillary.

When bulbs are maintained at different temperatures, let number of moles are $n_1$ and $n_2$ in bulbs at temperatures $T$ and $2 T$, respectively.

As gas is ideal,

$n_1=\frac{p V}{R T} \text { and } n_2=\frac{p V}{R(2 T)} \Rightarrow \frac{n_1}{n_2}=\frac{2}{1}$

Also, $\quad n_1+n_2=n$

$\Rightarrow \quad n_1=\frac{2 n}{3} \text { and } n_2=\frac{n}{3}$

Now, ratio of final mass of gas in bulb $2$ to the initial mass of gas in same bulb is

$=\frac{\text { Mass of gas finally in bulb } 2}{\text { Mass of gas initially in bulb } 2}$

$=\frac{\text { Number of moles finally present in bulb } 2}{\text { Number of moles initially present in bulb } 2}$

$=\frac{n / 3}{n / 2}=\frac{2}{3}$