The mass of glucose that should be dissolved in 50 g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1 g of urea in the same quantity of water is:

3 g

The correct answer is option 2. 3 g.

The formula you provided is the correct expression for the ratio of the lowering of vapor pressure for two different solutes in the same solvent.

\( \frac{{P_0 - P_s}}{{P_0}} = \frac{{n_2}}{{n_1}} = \frac{{w_2M_1}}{{w_1M_2}} \)

Where:
\( P_0 \) is the vapor pressure of the pure solvent,
\( P_s \) is the vapor pressure of the solution,
\( n_1 \) and \( n_2 \) are the number of moles of solute,
\( w_1 \) and \( w_2 \) are the masses of solute, and
\( M_1 \) and \( M_2 \) are the molar masses of solute.

We are given that the lowering of vapor pressure is the same for both cases.

For the first case, we have 1 g of urea (\( M_1 = 60 \, \text{g/mol} \)) dissolved in 50 g of water (\( w_1 = 50 \, \text{g} \)).

For the second case, we have an unknown mass of glucose (\( M_2 = 180 \, \text{g/mol} \)) dissolved in 50 g of water (\( w_2 = ? \)).

Using the given information and the formula, we can set up the equation:

\(\left(\frac{w_2M_1}{w_1M_2}\right)_{glucose} = \left(\frac{w_2M_1}{w_1M_2}\right)_{urea}\)

\( \frac{{1 \times 18}}{{50 \times 60}} = \frac{{w_2 \times 180}}{{50 \times 1}} \)

Simplifying the equation, we find:

\( w_2 = 3 \, \text{g} \)

Therefore, the mass of glucose that should be dissolved in 50 g of water to produce the same lowering of vapor pressure as 1 g of urea is 3 g, which corresponds to option (2).