Answer
$R = 4.59 \times 10^{-5} g/h$
Work Step by Step
$A = A_oe^{-\lambda t}$ and $\lambda = \frac{ln2}{T_{1/2}}$
* side note: $1 \space ci = 3.7 \times 10^{10} s^{-1} Ci^{-1}$ can be used later in conversion.
The activity after 1000 hours is
$A_{1000h} = A_oe^{-\frac{tln2}{T_{1/2}}}$
$A_{1000h} =(9.4 \times 10^{-6} Ci )( 3.7 \times 10^{10} s^{-1} Ci^{-1}) e^{-\frac{(1000h)ln2}{(45d)(24h/d)}}$
$A_{1000h} =1.8306 \times 10^5 Bq$
The activity is 84 Bq
current activity = $\frac{A}{A_{1000h}}$
current activity =$ \frac{84 Bq}{1.8306 \times 10^5 Bq}$
current activity =$ 4.5886 \times 10^{-4} $
Now at 84 Bq, the mass fraction within the activity is
$m_f = current \space activity \times m$
$m_f = 4.5886 \times 10^{-4} \times 100g$
$m_f = 4.59 \times 10^{-2} g$
This mass fraction is divided by 1000h because it is the total time elapsed in the test. The rate at which the piston rings lost their mass is
$R = \frac{m_f}{1000h}$
$R = \frac{4.59 \times 10^{-2} g}{1000h}$
$R = 4.59 \times 10^{-5} g/h$
