How many perfect cubes are there between 1 and 100000 which are divisible by 7?

A perfect cube that is divisible by 7 ,when that number contain a cube of 7 or a perfect 

multiple of 7's cube in following form , $$N=7^3\times k^3\ \left(where,\ k=1,2,3.....\right)$$

If we, put k=6 ,then the number become, $$N=7^3\times6^3=74088.$$

But , if we put k=7, then the number become, $$N=7^3\times7^3=117649.$$

117649 exceeds 100000 , So there are only 6 numbers(when, k=1,2,....,6) present between 1 and 100000 which are divisible by 7.

B is correct choice.