How to calculate what mass of halothane is inspired in one minute given the partial pressure and temperature?
Volumetric Flow Rate (VFR) is given to you. 
By applying ideal gas equation to the gas, in terms of volumetric flow rate, 
$$\frac{PV}{t} = \frac{nRT}{t}$$

And in an ideal gas mixture (the assumption), 
the partial pressure of a gas is what is exerts alone in same volume in same temperature as of the mixture.
$$\frac{P_\mathrm{gas}\cdot V_\mathrm{tot}}{t} = 
\frac{n_\mathrm{gas}\cdot R\cdot T_\mathrm{total}}{t}$$

amount of substance = mass of the compound/molar mass (or molecular weight of the compound)
$$n = \frac{m}{M}$$

$$\frac{P_\mathrm{gas}\cdot V_\mathrm{total}}{t} =
 \frac{(\frac{m_\mathrm{gas}}{M})\cdot R\cdot T_\mathrm{total}}{t}$$

Apply the values, mind the units:
\begin{align}
V &amp;=\pu{0.1 L}\\
T &amp;= (273+21)\pu{ K}\\
P_\mathrm{gas} &amp;= \pu{0.01 atm}\\
R &amp;=\pu{0.082057 L atm mol^-1K^-1}\\
M_\mathrm{gas} &amp;=\pu{197.4 g mol^-1}\\
t &amp;= \pu{1 min}\\
\end{align}

$$\frac{\pu{0.01 atm}\cdot\pu{0.1 L}}{\pu{1 min}} = 
\frac{
  \frac{m_\mathrm{gas}}{\pu{197.4 g mol^-1}}
  \cdot\pu{0.082057 L atm mol^-1 K^-1}\cdot\pu{294 K}}{\pu{1 min}}$$

For $m_\mathrm{gas}$ I can see the value in your multiple choice.