we have that
Josef’s sediment is 0.012 mm
Matthew’s deposit is 10^2 times as long as Josef’s
so
Matthew’s deposit=10^2*(0.012)=100*0.012=1.2 mm
therefore
The answer Part B is 1.2 mm
Part C
To obtain the answer to Part B, multiply the diameter of Josef’s sediment by the factor 10^2