Answer
$16\,\mu Ci$
Work Step by Step
Number of half-lives=$\frac{t}{t_{1/2}}=\frac{36\,h}{12\,h}=3$
The sample underwent 3 half-lives. That is, the initial activity is decreased by half three times. If $A_{0}$ is the initial activity, then
$A_{0}\times(\frac{1}{2})^{3}=2.0\,\mu Ci$
Or $A_{0}=2.0\,\mu Ci\times2^{3}=16\,\mu Ci$