#### Answer

Please see the work below.

#### Work Step by Step

We know that
$\frac{u_{H_2}}{u_{H_2se}}=(\frac{M_m(H_2Se)}{M_m(H_2)})^{\frac{1}{2}}$
We plug in the known values to obtain:
$\frac{u_{H_2}}{u_{H_2Se}}=(\frac{80.976}{2.016})^\frac{1}{2}=6.338$
Hence the ratio of the rates of effusion of $H_2$ and $H_2Se$ is 6.338 to 1.