Answer
$M_{Sample} = 0.657 g$
Work Step by Step
To calculate the mass of sample, we must use two equations which is
Equation 1
$N = \frac{R}{\lambda} = \frac{R}{ln2/T_{1/2}} = \frac{R (T_{1/2} )}{ln 2}$
Equation 2
$M_{Sample} = N \frac{M_K}{N_A}$
Substitute equation 1 into equation 2, we get
$M_{Sample} = [\frac{R (T_{1/2} )}{ln 2}] \frac{M_K}{N_A}$
From the question, we get the following information
$R = 1.70 \times 10^5 /s $
$T_{1/2} = 1.28 \times 10^9 y \times (3.15 \times 10^{7} s/y)$
$T_{1/2} = 4.0366 \times 10^{16} s$
$M_K = 40g/mol$
$N_A = 6.022 \times 10^{23}/mol $
Substitute all values into equation
$M_{Sample} = [\frac{(1.70 \times 10^5 /s ) (4.0366 \times 10^{16} s )}{0.693}] \frac{40g/mol}{6.022 \times 10^{23}/mol}$
$M_{Sample} = (9.90 \times 10^{21} ) (6.64 \times 10^{-23})$
$M_{Sample} = 0.657 g$
