$PV = \frac{m}{M} RT$ (for m grams of gas) $\quad...\left(i\right)$
$P' V = \frac{m'}{M} RT'\quad... \left(ii\right)$
Dividing equation $\left(ii\right)$ by $\left(i\right)$ we get
$\frac{P'}{P} = \frac{m'}{m} \frac{T'}{T}$
$m' = \frac{P'}{P} \times \frac{T}{T'} \times m = \left(\frac{P /2}{P}\right) \times \frac{400}{300} \times 6 = 4\,g$
$\therefore\quad$ Mass of oxygen leaked, $\Delta m = m -m ' = 6 - 4 = 2 \,g$