At 400K, the root mean square (rms) speed of gas X (molecular mass=40) is equal to most probable speed of gas Y at 60K. The molecular mass of the gas Y is:

Most probable speed of a gas is the speed possessed by the maximum number of gas molecules.

The formula for root mean square speed is $$V_{rms}=\sqrt{{\displaystyle \frac{3RT}{M}}}$$
And, the formula for most probable speed is $$V_{mps}=\sqrt{{\displaystyle \frac{2RT}{M}}}$$
Here, R= gas constant, T= absolute temperature and M= molar mass of the gas.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{rms}} \mathrm{of} \mathrm{X} \mathrm{gas}=\sqrt{\frac{3\mathrm{RTx}}{\mathrm{Mx}}} \\ {\text{ V}}_{\mathrm{mps}}\text{ of Y gas=}\sqrt{\frac{2\mathrm{RTy}}{\mathrm{My}}} \end{gathered}$$

According to the given question:

$$\begin{gathered} \left({\mathrm{V}}_{\mathrm{rms}} \mathrm{of} \mathrm{X} \mathrm{gas}\right)=\left({\text{ V}}_{\mathrm{mps}} \mathrm{of} \mathrm{Y} \mathrm{gas}\right) \\ \sqrt{\frac{3{\mathrm{RT}}_{\mathrm{X}}}{{\mathrm{M}}_{\mathrm{X}}}}=\sqrt{\frac{2{\mathrm{RT}}_{\mathrm{Y}}}{\mathrm{My}}} (\mathrm{squaring} \mathrm{both} \mathrm{sides}) \\ \frac{3{\mathrm{RT}}_{\mathrm{X}}}{{\mathrm{M}}_{\mathrm{X}}}=\frac{2{\mathrm{RT}}_{\mathrm{Y}}}{{\mathrm{M}}_{\mathrm{y}}} \\ {\mathrm{M}}_{\mathrm{y}}=\frac{2{\mathrm{T}}_{\mathrm{Y}}{\mathrm{M}}_{\mathrm{X}}}{3{\mathrm{T}}_{\mathrm{X}}} \\ {\mathrm{M}}_{\mathrm{y}}=\frac{2\times 60\times 40}{3\times 400 }=4 \end{gathered}$$