Finding the density of nitrogen at 0 degrees Celsius and 1 atmosphere
Using the ideal gas law
\begin{align}
  pV &amp;= nRT, &amp;
  \frac nV &amp;= \frac{p}{RT},
\end{align}
where
\begin{align}
  T &amp;= \pu{273 K},\\
  p &amp;= \pu{1 atm},\\
  R &amp;= \pu{0.0821 L atm mol-1 K-1},\\
  M(\ce{N2}) &amp;= \pu{28.0 g mol-1},
\end{align}
we get
\begin{align}
  \frac{n}{V} &amp;= 
    \frac{\pu{1 atm}}{\pu{0.0821 L atm//mol K} \times\pu{273 K}},\\
  \frac{n}{V} &amp;= \pu{0.0446 mol//L},
\end{align}
and finally
$$
\rho(\ce{N2}) 
  = \pu{0.0446  mol//L} \times \pu{28  g//mol} 
  = \pu{1.2  g//L}.
$$
Therefore answer (d) is correct.