what rate of interest compounded continuously is required to double an investment in 3 years?

Ln(2) / 3 ==23.105% compounded continuously will double your money in 3 years.

Let r% be such rate of interest.

\(P\displaystyle\lim_{n\to\infty}\left(1 + \dfrac{r\%}n\right)^{3n} = 2P\\ e^{3r\%} = 2\\ 3r\% = \ln(2)\\ r\% = \dfrac{\ln(2)}3 \approx 0.231\)

The rate must be at least 23.1%.