Bell metal contains $80\%$ copper. The mass of Bell metal which contains $1.5\times {10}^{20}$ atoms of copper is 

We know that $6.023x{10}^{23}$ atoms will be there in $64g$ of copper 

(Molecular Weight of copper = $64g$).

Then, $1.5x{10}^{20}$ atoms will contain $6.023x{10}^{23}/64x1.5x10{ }^{20}$ g of copper.

That is, $16x10{ }^{-3}$ g of copper will be present in $1.5x10{ }^{20}$ atoms.

Now, it is given that bell metal contains $80\%$ of copper that is $100g$ of bell metal contains 

$80g$ of copper.

$80g$ of copper is present in $100g$ of bell metal then, $16x10{ }^{-3}$ g of copper will be  

present in $100gx16x{10}^{-3}g/80g$

That is, $20x{10}^{-3}$ g or $20 mg$

Therefore, Option (B) is correct.