Answer
$3.46897\times 10^{18}\;\rm s$
Work Step by Step
We know that the half-life for a decay process is given by
$$R=\lambda N=\dfrac{\ln 2}{T_{\frac{1}{2}}}N_{\rm ^{147}_{62}Sm}$$
Solving for $T_{\frac{1}{2}}$;
$$T_{\frac{1}{2}}=\dfrac{\ln 2}{R}N_{\rm ^{147}_{62}Sm}\tag 1$$
Now we need to find $N_{\rm ^{147}_{62}Sm}$ and we know that its amount in a natural sample is about $15\%$. So that
$$N_{\rm ^{147}_{62}Sm}=0.15 N_{natural }$$
Plugging the known;
$$N_{\rm ^{147}_{62}Sm}=0.15\left(\dfrac{1\times 6.02\times 10^{23}}{150.36}\right)$$
where 150.36 is the samarium atomic mass in g/mole and $6.02\times 10^{23}$ is Avogadro's number which is the number of atoms in one mole.
$$N_{\rm ^{147}_{62}Sm}=\bf 6.0056\times 10^{20}\rm nuclei$$
Plugging into (1) and plug the known as well;
$$T_{\frac{1}{2}}=\dfrac{\ln 2}{120} \times 6.0056\times 10^{20} $$
$$T_{\frac{1}{2}}= \color{red}{\bf 3.46897\times 10^{18}}\;\rm s=1.09925\times 10^{11}\;\rm year$$