Answer
$64\,\mu Ci$
Work Step by Step
Number of half-lives=
$\frac{t}{t_{1/2}}=\frac{27\,days}{4.5\,days}=6$
The sample underwent 6 half-lives.
That is, the initial activity is decreased by half 6 times. If $A_{0}$ is the initial activity, then
$A_{0}\times(\frac{1}{2})^{6}=1.0\,\mu Ci$
Or $A_{0}=1.0\,\mu Ci \times 2^{6}=64\,\mu Ci$