First, note that, since the digit sum of 19683 is a multiple of 9, 19683 itself is a multiple of 9. Second, since 19683 is odd, its cube root will be odd as well. Therefore, the cube root of 19683 will be an odd multiple of 3.
To find 19683¹ᐟ³, will take the base 10 log of 19683, divide by 3, and raise the result to the 10th power, approximating where convenient.
To this end, it will be helpful to have approximate values for lg 2 and lg 5. Since 10³ ≈ 1024 = 2¹⁰ = 10^(lg[2]·10), we have 3 ≈ lg[2]·10. Solving, we find that lg 2 ≈ 0.3. Next, since 10 = 2·5, we have lg[10] = 1 = lg[2·5] = lg 2 + lg 5, so lg 5 = 1 − lg 2 ≈ 0.7.
Using those approximations, we calculate:
lg[19683¹ᐟ³] ≈ lg[20000¹ᐟ³] = lg[2·10000]/3 = (lg 2 + lg 10000)/3 ≈ (0.3 + 4)/3 ≈ 1.43.
Since 25 = 5·5, we have lg 25 = lg 5 + lg 5 ≈ 1.4. This means we're looking for an odd multiple of 3 near 25. The closest number that fits the bill is 27. Cubing, we find that 27·27·27 = 19683, so we are done.