At STP the density of a gas `X` is three times that of gas `Y` while molecule mass of gas Y is twice that of X. The ratio of pressures of X and Y will be:

Explanation

To solve the problem, we need to find the ratio of pressures of gas X and gas Y based on the given information about their densities and molecular masses.

### Step-by-Step Solution:

1. **Identify the Given Relationships:**
- The density of gas X (denoted as \(d_X\)) is three times that of gas Y (denoted as \(d_Y\)):
\[ d_X = 3 d_Y \]
- The molecular mass of gas Y (denoted as \(M_Y\)) is twice that of gas X (denoted as \(M_X\)):
\[ M_Y = 2 M_X \]

2. **Use the Formula for Pressure:**
- The pressure of a gas can be expressed in terms of its density and molecular mass using the formula:
\[ P = \frac{dRT}{M} \]
- Here, \(P\) is the pressure, \(d\) is the density, \(R\) is the universal gas constant, \(T\) is the temperature, and \(M\) is the molecular mass.

3. **Write the Expressions for Pressures of X and Y:**
- For gas X:
\[ P_X = \frac{d_X RT}{M_X} \]
- For gas Y:
\[ P_Y = \frac{d_Y RT}{M_Y} \]

4. **Find the Ratio of Pressures \(P_X\) and \(P_Y\):**
- The ratio of pressures can be expressed as:
\[ \frac{P_X}{P_Y} = \frac{\frac{d_X RT}{M_X}}{\frac{d_Y RT}{M_Y}} = \frac{d_X M_Y}{d_Y M_X} \]
- The \(RT\) terms cancel out.

5. **Substitute the Known Values:**
- Substitute \(d_X = 3 d_Y\) and \(M_Y = 2 M_X\) into the ratio:
\[ \frac{P_X}{P_Y} = \frac{(3 d_Y)(2 M_X)}{d_Y M_X} \]

6. **Simplify the Expression:**
- Cancel \(d_Y\) and \(M_X\):
\[ \frac{P_X}{P_Y} = \frac{3 \cdot 2}{1} = 6 \]

7. **Final Ratio:**
- Therefore, the ratio of pressures \(P_X\) to \(P_Y\) is:
\[ P_X : P_Y = 6 : 1 \]

### Conclusion:
The ratio of pressures of gas X to gas Y is \(6 : 1\).