a
Consider $STP$ conditions:
$(1)$ $1 \; mol =22.4 L$ of gas; $1 \; mol$ of Nitrogen gas $=14 g$
$(2)$ Therefore, $22.4 \; L$ Nitrogen gas $=14 \; g \rightarrow 2 \; L$ Nitrogen $=1.25 \; g$
$(3)$ This is at a standard pressure of $1 \; atm$. So, $22.4 \; atm$ pressure corresponds to $22.4 \times 1.25=28 \; g$