Oh,I see,It's identical to the earth's calculation but as far as I am concerned,question (b) and (a) should be exchanged both place,which means we should know the radius of Mercury first.
There your answers go:
we assume the mass of satellite is" m "；distance of that satellite above the surface of Mercury is " H" Its Angular velocity is "ω" ,cycle time is T.Mercury 's radius is "R"
According to the Law of universal gravitation:
GmM/(r+H)²=mω²（r+H）=m4π²/T²（r+H）（only reason is that we can hereby put the cycle time in ues)
M =4π²(R+H)³/GT² ≈3.3*10 ^23 kg
the volume of a spherical is 4πR³/3
M=ρv=ρ*4πR³/3
the density ρ=3M/4πR³≈5.430g/cm³=5430kg /m³
the density of the earth is 5518kg/m³,which is a little larger than Mercury's.In addition,the radius of our planet is nearly three times bigger than that of Mercury.So,Mercury's size is about 27 times smaller than our planet.