The half-life of Au ^ 198 is 2 . 7 days. The activity of 1 . 50mg...

The activity will be

$A = \lambdaN$

Now,

$\lambda = \dfrac{\ln \left(2\right)}{{t}_{\dfrac{1}{2}}}$ and $N = \dfrac{1 . 5 \times {10}^{- 3}}{198} \times 6 \times {10}^{23}$

Therefore, activity in $Ci$,

$A = \dfrac{\ln 2}{{t}_{\dfrac{1}{2}}} \times \dfrac{1 . 5 \times {10}^{- 3}}{198} \times \dfrac{6 \times {10}^{23}}{3 . 7 \times {10}^{- 10}} = \dfrac{\ln 2}{2 . 7 \times 24 \times 60 \times 60} \times \dfrac{1 . 5 \times {10}^{- 3}}{198} \times \dfrac{6 \times {10}^{23}}{3 . 7 \times {10}^{- 10}} = 366Ci \approx 357Ci$