What could be the whole number, to which when 135,002 is added to its cube, we get the cube of a slightly larger whole number?
The difference of two cube numbers is even, so either both are even or both are odd, and thus the same applies to their bases. Then there is an integer between them (their arithmetic mean), let this be $a$, so the two numbers can be written in the form $a-k$ and $a+k$, where $k$ is a positive integer. According to the conditions of the problem,

$$
(a+k)^{3}-(a-k)^{3}=6 a^{2} k+2 k^{3}=2 k\left(3 a^{2}+k^{2}\right)=135002
$$

Simplifying and factoring the number on the right side,

$$
k\left(3 a^{2}+k^{2}\right)=7 \cdot 9643
$$

where the second factor is a prime number, as it is not divisible by any prime number less than or equal to its square root, i.e., any of those under 98. Thus, $k$, as a divisor of the right side, can only be 1 or 7.

In the case of $k=7$, $(9843-49) / 3=3198$ is not a square number, while for $k=1$, $a^{2}=22500, a= \pm 150$, so the numbers in question can be 149 and -151.