Answer
${\bf 0.371}$
Work Step by Step
We have a 100 g sample that contains two radioactive isotopes:
1. $ ^{131}\text{Ba} $ with a half-life of 12 days.
2. $ ^{47}\text{Ca} $ with a half-life of 4.5 days.
Initially, there are twice as many $ ^{47}\text{Ca} $ atoms as $ ^{131}\text{Ba} $ atoms.
$$ (N_0)_{\text{Ca}} = 2 (N_0)_{\text{Ba}} \tag 1$$
We need to find the ratio of calcium atoms to barium atoms after 2.5 weeks (or 17.5 days).
The number of atoms remaining at any time $ t $ is given by
$$
N = N_0 e^{-t / \tau}\tag 2
$$
where $ \tau $ is the mean lifetime where $ t_{1/2} $ by $ \tau = \frac{t_{1/2}}{\ln (2)} $
The ratio of $ ^{47}\text{Ca} $ atoms to $ ^{131}\text{Ba} $ atoms after $ t = 17.5 $ days is given by (2);
$$
\frac{N_{\text{Ca}}}{N_{\text{Ba}}} = \frac{(N_0)_{\text{Ca}} e^{-\ln (2)t /(t_{1/2})_{\text{Ca}}}}{(N_0)_{\text{Ba}} e^{-\ln (2)t /(t_{1/2})_{\text{Ba}}}}
$$
Plug form (1);
$$
\frac{N_{\text{Ca}}}{N_{\text{Ba}}} = \frac{2(\color{red}{\bf\not} N_0)_{\text{Ba}} e^{-\ln (2)t /(t_{1/2})_{\text{Ca}}}}{(\color{red}{\bf\not} N_0)_{\text{Ba}} e^{-\ln (2)t /(t_{1/2})_{\text{Ba}}}}
$$
Plug the known;
$$
\frac{N_{\text{Ca}}}{N_{\text{Ba}}} = \frac{ 2e^{-\ln (2)(17.5) /(4.5) }}{ e^{-\ln (2)(17.5) / (12)}}=\color{red}{\bf 0.371}
$$
