Finding the amount of nitrous oxide produced in the reaction between lead sulfide and nitric acid
The first step is to calculate the number of moles of $\ce{PbS}$:
$$n(\ce{PbS}) = \frac{\pu{4.7 g}}{\pu{239 g mol-1}} = \pu{0.0197 mol}$$
The next step is to calculate how many moles of $\ce{NO}$ will be produced based on the stoichiometry of the reaction. As stated in a comment, it looks like this is the step that was left out of your calculations. Because only $\pu{2 mol}$ of $\ce{NO}$ are produced for every $\pu{3 mol}$ of $\ce{PbS}$, only $2/3$ of $\pu{0.0197 mol}$, or $\pu{0.0131 mol}$ $\ce{NO}$ will be formed.
At this point we know all of the variables required to invoke the ideal gas law:
$$
\begin{align}
V(\ce{NO}) &amp;= \frac{nRT}{p}\\
 &amp;= \frac{\pu{0.0131 mol} \times \pu{0.0821 L atm mol-1 K-1} \times \pu{300K}}{\pu{1.10 atm}}\\   
 &amp;= \pu{0.29 L}
\end{align}
$$
And this gives us the correct answer of a. Again, as your answer was off by a factor of $2/3$, it is pretty clear that the second step above regarding the stoichiometry of the reaction is the only step that you left out of your calculations.