Answer
$1.3\mu Ci$
Work Step by Step
We know that
$\lambda=\frac{\ln 2}{T_{\frac{1}{2}}}$
$\lambda=\frac{0.693}{6.05h}=0.115/h$
$Number\space of\space nuclei \space N=\frac{R}{\lambda}$
$N=\frac{1.5\times 10^{-6}Ci}{0.115/h}$
$N=(1.5\times 10^{-6}Ci)(\frac{1}{0.155/h})(\frac{3600s}{h})(\frac{3.7\times 10^{10}decay/s}{1Ci})=1.74\times 10^9nuclei$
Now the required activity can be determined as
$R=\lambda N_{\circ}e^{-\lambda}t$
We plug in the known values to obtain:
$R=(0.115/h)(1.744\times 10^9)e^{-\frac{0.115/h}{2.05h}}(\frac{1h}{3600s})(\frac{1ci}{3.7\times 10^{10}decay/s})$
$R=1.3\mu Ci$
