Answer:
The correct answer is option B
Explanation:

Given values,
Molarity of 
Volume of solution, 
Molecular weight of 
Substituting this values in Molarity formula, we get

Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag 6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ → ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag
∅ | 63.5 g Cu | 1 mol Cu | 1 mol Ag 63.5
There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.