Answer
$64\,\mu Ci$
Work Step by Step
Number of half-lives=
$\frac{t}{t_{1/2}}=\frac{130\,days}{32.5\,days}=4$
The sample underwent 4 half-lives.
That is, the initial activity is decreased by half four times. If $A_{0}$ is the initial activity, then
$A_{0}\times(\frac{1}{2})^{4}=4.0\,\mu Ci$
Or $A_{0}=4.0\,\mu Ci\times2^{4}=64\,\mu Ci$
