If there is a trend that $\frac{1}{3}$rd of a population of a community has migrated every year from one place to some other place, what is the leftover population of that community after the sixth year, if there is no further growth in the population during this period ?

The correct option is $\frac{64}{729}$rd part of the population.

Step-by-step Explanation:

1. Understand the Fraction of Population Remaining Each Year:
Let the initial population of the community be represented as $1$ (or $P$).
It is given that $\frac{1}{3}$rd of the population migrates away every year.
Therefore, the fraction of the population that stays or remains at the end of any given year is:
$1-\frac{1}{3}=\frac{2}{3}$

2. Calculate the Remaining Population After 6 Years:
Since there is no further growth in the population during this period, the remaining population decreases by a factor of $\frac{2}{3}$ each year for $6$ consecutive years.

After the 1st year, leftover population = $\frac{2}{3}$
After the 2nd year, leftover population = ${\left(\frac{2}{3}\right)}^2$
After the 6th year, leftover population = ${\left(\frac{2}{3}\right)}^6$

3. Evaluate the Power:
${\left(\frac{2}{3}\right)}^6=\frac{2^6}{3^6}$

Calculating the numerator and denominator separately:
$2^6=2\times 2\times 2\times 2\times 2\times 2=64$
$3^6=3\times 3\times 3\times 3\times 3\times 3=729$

Thus, the leftover population after the sixth year is $\frac{64}{729}$rd part of the population.