#(g_e)/(g_m)=6.05#

The data given in the question is way out. The mass of the moon is 1/81.3 of the earth so I'll work through with that.

I'll also use the radius of the moon to be 0.273 x the radius of the earth.

We'll compare an object of mass #m# on the surface of the earth with an object of mass #m# on the surface of the moon.

Newton says that the force of attraction between 2 objects #M# and #m# is:

#F=(GmM)/(R^(2)#

For this problem:

#F_e# = force of attraction on the earth

#F_m# = force of attraction on the moon

#R_e# = radius earth

#R_m# = radius moon

#m# = mass object

So:

#F_e=(GmM_e)/(R_e^(2)# Eqn #color(red)((1))#

#F_m=(GmM_m)/(R_m^(2))# Eqn#color(red)((2))#

We know that:

#R_m=0.273R_e#

and:

#M_m=M_e/81.3#

So we can substitute these values into #color(red)((2))# #rArr#

#F_m=(Gm[(M_e)/(81.3)])/((0.273R_e)^(2))#

#F_m=(Gm[(M_e)/(81.3)])/(0.0745R_e^(2))# Eqn #color(red)((3))#

Now divide #color(red)((1))# by #color(red)((3))# #rArr#

#(F_e)/(F_m)=(GmM_e)/(R_e^(2))xx(0.0745R_e^(2))/(Gm[[M_e)/(81.3)]#

This cancels down to:

#(F_e)/(F_m)=(0.0745)/((1)/(81.3))=6.05#

Since #F=ma#:

#(mg_e)/(mg_m)=6.05#

So:

#(g_e)/(g_m)=6.05#

This agrees with the generally accepted value of the acceleration due to gravity of the moon being about 1/6 of that of the earth.

You can substitute the values given in the question if you want to but I would check their source.