What mass of silver ag which has an atomic mass of 107.87 amu contains the same number of atoms contained in 10.0g of boron b which has an atomic mass of 10.81 amu?

To contain the same number of atoms also mean to contain the same number of moles. So let us say that X is the mass of Silver Ag required, so that:

X / 107.87 = 10 / 10.81

X = 99.79 g