Answer
$5.61\times10^{-10}kg$.
Work Step by Step
Assume that the activity is constant for a short period. Apply the definition of the curie.
$$(2.50\times10^{-6}Ci)\frac{3.70\times10^{10}decays/s}{1\;Ci}=9.25\times10^4 decays/s$$
$$9.25\times10^4 decays/s=\frac{\Delta N}{\Delta t}=N\frac{0.693}{T_{1/2}}$$
$$N=(9.25\times10^4 decays/s)\frac{(5730yr)(3.156\times10^7 s/yr)}{0.693}$$
$$=2.414\times10^{16}nuclei$$
Let’s find the mass.
$$2.414\times10^{16}nuclei (\frac{0.0140kg}{6.02\times10^{23}nuclei})=5.61\times10^{-10}kg$$
