The solution of the differential equation is $$x(t)=x_0e^{-kt}.$$
To say that the half-life is $h$ is to say that $x(h)=\frac{x_0}{2}$. Substituting in the formula for $x(t)$, we get $$x_0e^{-kh}=\frac{x_0}{2}.$$
Cancel the $x_0$, and take reciprocals (this part is not strictly necessary). We get $$e^{kh}=2.$$ Take the natural logarithm, which we will call $\ln$. We get $$kh=\ln 2\quad\text{and therefore} \quad k=\frac{\ln 2}{h}.$$ Now use the given value $700\times 10^6$ of the half-life. I get that $k$ is approximately $9.902\times 10^{-10}$.