1. Use freezing point depression relation

For dilute solutions, $\Delta T_f = iK_f m$

Since both are aqueous solutions, the solvent is same, so $K_f$ is same. Assuming both solutes are non-electrolytes, $i=1$.

Given that the freezing points are equal, their depressions in freezing point are equal: $\Delta T_{f1} = \Delta T_{f2}$ So, $m_1 = m_2$

2. Interpret percentage composition

A 4% aqueous solution of $X$ means:

• $4$ g of $X$ in $100$ g of solution
• Hence, water $= 96$ g $= 0.096$ kg

Moles of $X$: $\frac{4}{A}$

Molality of $X$: $m_1 = \frac{4/A}{0.096}$

3. For 12% aqueous solution of $Y$

A 12% aqueous solution of $Y$ means:

• $12$ g of $Y$ in $100$ g of solution
• Hence, water $= 88$ g $= 0.088$ kg

Let molecular weight of $Y = M$.

Moles of $Y$: $\frac{12}{M}$

Molality of $Y$: $m_2 = \frac{12/M}{0.088}$

4. Equate molalities

$\frac{4/A}{0.096} = \frac{12/M}{0.088}$

$\frac{4}{0.096A} = \frac{12}{0.088M}$

Cross-multiplying: $4 \times 0.088 M = 12 \times 0.096 A$

$0.352M = 1.152A$

$M = \frac{1.152}{0.352}A$

$M = 3.27A \approx 3A$

5. Choose nearest option

Thus the molecular weight of $Y$ is approximately: $\boxed{3A}$

So, the correct option is C.