How to calculate the volume or mass of carbon dioxide gas absorbed by a calcium hydroxide solution?
(a) It consumes $\pu{4.4 L}$ of carbon dioxide gas


(wrong)

You've correctly calculated the volume of carbon dioxide $V(\ce{CO2})$ via the molar volume of ideal gas $V_\mathrm{m} = \pu{24.790 L mol-3}$ at SATP (Standard Ambient Temperature and Pressure):

$$V(\ce{CO2}) = nV_\mathrm{m} = \pu{0.1 mol}\cdot\pu{24.790 L mol-3} \approx \pu{2.5 L}$$

I personally wouldn't recommend to remember molar volumes at STP, NTP, SATP etc. as the numbers are close, but still different, and it brings up a lot of confusion.
Instead, only remember the ideal gas constant $R$, and remember it well as it's being used ubiquitously.
Pretending you forgot the molar volume for these conditions, you can use ideal gas law directly:

$$V(\ce{CO2}) = \frac{nRT}{p} = \frac{\pu{0.1 mol}\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{298 K}}{\pu{1e5 Pa}} \approx\pu{2.5 L}$$

Here, keep in mind that $\pu{1 Pa} = \pu{1 J m-3}$ and $\pu{1 m3} = \pu{1e3 L}$.


  (b) It produces $\pu{24.5 L}$ of liquid water


(wrong)

Liquid water and SATP both imply that the density of water $ρ(\ce{H2O})\approx\pu{1 g mL-1}$, and its volume can be found via molecular mass $M(\ce{H2O})$:

$$V(\ce{H2O}) = \frac{m(\ce{H2O})}{ρ(\ce{H2O})} = \frac{nM(\ce{H2O})}{ρ(\ce{H2O})} = \frac{\pu{0.1 mol}\cdot\pu{18.02 g mol-1}}{\pu{1 g mL-1}} \approx \pu{1.8 mL}$$ 


  (c) It consumes $\pu{0.5 mol}$ of carbon dioxide


(wrong)

The simplest out of four.
As you've already ruled out, there are equimolar amounts of both reactants and products according to the stoichiometry, so $n(\ce{CO2}) = n(\ce{Ca(OH)2} = \pu{0.1 mol})$:

$$\ce{Ca(OH)2 (aq) + CO2 (g) -&gt; CaCO3 (s) + H2O (l)}$$


  (d) It produces $\pu{10.0 g}$ of solid calcium carbonate


(correct)

Mass of calcium carbonate $m(\ce{CaCO3})$ can be found trivially from its molar mass $M(\ce{CaCO3})$:

$$m(\ce{CaCO3}) = nM(\ce{CaCO3}) = \pu{0.1 mol}\cdot\pu{100.09 g mol-1} \approx \pu{10.0 g}$$