The half- life of $A{u}^{198}$ is 2.7 days. The activity of 1.50 mg of $A{u}^{198}$  , if its atomic weight is 198 g $mo{l}^{-1}$ , is…….   ($N_A =6\times {10}^{23}/mol$).

Initial activity  $=A_0=\lambda N_0.$$=\frac{\ln 2}{T_{1/2}}\times \frac{1.5\times {10}^{-3}}{198}\times 6.023\times {10}^{23}$ 
$=\frac{\ln 2}{2.7\times 3600\times 24}\times \frac{1.5\times {10}^{-3}}{198}\times \frac{6.023\times {10}^{23}}{3.7\times {10}^{10}}Ci$ 
= 366.4 Ci
Closest answer is 357 Ci so option Key 3 is correct.