Answer
(a) greater than
(b) $2.07Km/s$
Work Step by Step
(a) We know that the rms speed is inversely proportional to the mass. Thus, the rms speed of $H_2O$ is greater than $O_2$ as $H_2O$ is has less mass as compared to that of $O_2$.
(b) We can find the required rms speed of $H_2O$ as follows:
$v_{rms \space H_2O}=v_{rms\space O_2}\sqrt{\frac{M_{O_2}}{M_{H_2O}}}$
We plug in the known values to obtain:
$v_{rms}=(1550m/s)\sqrt{\frac{32.0g/mol}{18.015g/mol}}$
$v_{rms}=(1550m/s)(1.33)$
$v_{rms}=2065.8m/s=2.07Km/s$
