The solubility product constants of Ag$$_2$$CrO$$_4$$ and AgBr are 32x and 4y respectively at 298 K. The value of $$\left(\frac{\text{molarity of Ag}_2\text{CrO}_4}{\text{molarity of AgBr}}\right)$$ can be expressed as :

Let the molar solubility (molarity of the saturated solution) of $$Ag_2CrO_4$$ be $$S_1$$ M.

$$Ag_2CrO_4 \; (s) \;\rightleftharpoons\; 2\,Ag^+ (aq) + CrO_4^{2-} (aq)$$

At equilibrium: $$[Ag^+] = 2S_1$$ and $$[CrO_4^{2-}] = S_1$$.

For a salt $$A_mB_n$$ that dissolves as $$mA^{n+} + nB^{m-}$$, the solubility product expression is $$K_{sp} = [A^{n+}]^m [B^{m-}]^n$$. 

Applying this rule:

$$K_{sp}(Ag_2CrO_4) = [Ag^+]^2[CrO_4^{2-}] = (2S_1)^2(S_1)=4S_1^{\,3} \quad -(1)$$

The question states $$K_{sp}(Ag_2CrO_4)=32x$$, so from $$(1)$$

$$4S_1^{\,3}=32x \;\;\Longrightarrow\;\; S_1^{\,3}=8x \;\;\Longrightarrow\;\; S_1=2x^{1/3} \quad -(2)$$

Let the molar solubility of $$AgBr$$ be $$S_2$$ M.

$$AgBr \; (s) \;\rightleftharpoons\; Ag^+ (aq) + Br^- (aq)$$

At equilibrium: $$[Ag^+]=S_2$$ and $$[Br^-]=S_2$$.

For a salt that dissociates 1:1, $$K_{sp}=S_2^{\,2} \quad -(3)$$

The question gives $$K_{sp}(AgBr)=4y$$, hence from $$(3)$$

$$S_2^{\,2}=4y \;\;\Longrightarrow\;\; S_2=2y^{1/2} \quad -(4)$$

The required ratio is

$$\frac{\text{molarity of } Ag_2CrO_4}{\text{molarity of } AgBr}=\frac{S_1}{S_2}=\frac{2x^{1/3}}{2y^{1/2}}=\frac{\sqrt[3]{x}}{\sqrt{y}}$$

Therefore, the correct choice is
Option D i.e $$\frac{\sqrt[3]{x}}{\sqrt{y}}$$