How to find the rate of effusion given rates?
HINT:

This question seems to be a simple application of Graham's Law (assuming you know Graham's Law), i.e the rate of effusion of a gas is inversely proportional to the square root of the mass of their particles. This empirical law can be formulated as follows:

$$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{\text{M}_2}{\text{M}_1}}$$

Solution: 


   I don't know what the OP meant by $\ce{NH}$, I assume the gas in question is $\ce{NH3}$ with rate of effusion ($R_1$) = 0.050 $\mathrm{mol \min^{-1}}$.

 $R_2 = \frac{R_1}{2}$ and $M_1 = 17.031$, plugging all of this in

 $M_2 = M_1\cdot \left (\frac{R_1}{R_2} \right )^2 = 4\cdot M_1 = 68.124$