Answer
1975
Work Step by Step
$\frac{A}{A_{0}}=\frac{17}{100}=0.17$ where $A_{0}$ is the original amount and $A$ is the amount after time $t$.
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{12.3\,y}=0.05634\,y^{-1}$
Recall that $\ln(\frac{A}{A_{0}})=-kt$
$\implies \ln(0.17)=-1.77=-(0.05634\,y^{-1})(t)$
$\implies t=\frac{-1.77}{-0.05634\,y^{-1}}=31.4\,y$
The watch can be used until (1944+31)=1975