Alcohol fermentation is the formation of alcohol from sugar (or glucose). Yeast, when under anaerobic conditions, convert one glucose molecule $(\ce{C6H12O6})$ to two molecules of pyruvic acid $(\ce{CH3C(=O)CO2H})$ via the glycolysis pathways, then go one step farther, converting each pyruvic acid into ethanol $(\ce{CH3CH2OH})$ and carbon dioxide $(\ce{CO2})$ via acetaldehyde:
Thus, each glucose molecule would form two ethanol molecules and two $\ce{CO2}$ molecules:
$$\ce{C6H12O6 ->[yeast] 2C2H5OH + 2 CO2}$$
Let's look at what we have in the flask: $\pu{50 mL}$ of $20\% (w/v)$ glucose solution, density of which is $\pu{1.08 g mL-1}$. Thus, mass of $\pu{50 mL}$ solution is:
$$\pu{50 mL} \times \pu{1.08 g/mL} = \pu{54 g}$$
Since $\pu{10 g}$ of this mass is the mass of glucose, the mass of water in the $\pu{50 mL}$ solution is $\pu{44 g}$. Assuming density of water is $\pu{1.0 g mL-1}$ at $\pu{30 ^\circ C}$, then the volume of water at this temperature is $\pu{44 mL}$ (Note: This assumption is justified by the facts that neither densities of solutions are given in particular temperatures nor the $\pu{50 mL}$ of solution is measured in same temperature).
If the solution have gone to 100% completion, none of $\pu{10 g}$ of glucose $(MW = \pu{108.16 g mol-1})$ will be remaining in the solution and as a result, the volume would decrease to $\pu{44 mL}$ of water. Let's calculate the mass ($m$) of ethanol $(MW = \pu{46.07 g mol-1})$ would be produced by this fermentation:
$$m = \frac{\pu{10 g}}{\pu{180.16 g mol-1}} \times 2 \times \pu{46.07 g mol-1} = \pu{5.11 g}$$
Therefore maximum volume of ethanol would produce by fermentation is: $\frac{\pu{5.11 g}}{\pu{0.789 g mL-1}} = \pu{6.48 mL}$. Thus, assuming ethanol and water are additive, maximum volume of the solution in the flask is: $\pu{6.48 mL} + \pu{44 mL} = \pu{50.48 mL}$.
Thus, this reaction would increase the volume of the solution only by $\pu{0.48 mL}$, if the reaction went to 100% conversion, so that the data given is incorrect (can't increase by $\pu{5.00 mL}$ as given).
Thus, assuming 100% conversion, we can calculate the amount of $(\ce{CO2})$ produced ($m_1$) in the flask:
$$m_1 = \frac{\pu{10 g}}{\pu{180.16 g mol-1}} \times 2 = \pu{0.111 mol}$$
Assuming no $(\ce{CO2})$ would dissolve in aqueous solution (this is actually incorrect assumption but it was instructed in the question) and $(\ce{CO2})$ behaves as an ideal gas, the partial pressure inserted by the $(\ce{CO2})$ gas ($p$) is:
$$p = \frac{nRT}{v} = \frac{\pu{0.111 mol} \times \pu{0.0821 atm L mol-1 K-1} \times \pu{303 K}}{\pu{(0.125-0.05048) L}} = \pu{37.1 atm}$$
