Answer:
46.2 years
Explanation:
You recognize that the solution to the differential equation will be ...
m(t) = c·e^(-0.015t) . . . . . . .for some c dependent upon initial conditions
and that the half-life will be the answer to ...
ln(1/2) = -0.015t . . . . . . . . . use c=1, m=1/2, and take the natural log
t = ln(1/2)/-0.015 ≈ 46.2
The half-life will be about 46.2 years.