Let Mr. X's age last year be the square of a number, say $n^2$. Then his age next year will be the cube of a number, say $m^3$.

Let Mr. X's current age be $x$.

So, from the information:

• Last year, $x - 1 = n^2$
• Next year, $x + 1 = m^3$

Thus, we have the equation: $x - 1 = n^2$ and $x + 1 = m^3$ By subtracting the first equation from the second: $(x + 1) - (x - 1) = m^3 - n^2$ $2 = m^3 - n^2$

Now, we need to find integer values of $m$ and $n$ such that $m^3 - n^2 = 2$.

Step 1: Try different values of $n$ and $m$

• For $n = 1$: $n^2 = 1$ $m^3 = 3$ (no integer cube of 3)

• For $n = 2$: $n^2 = 4$ $m^3 = 6$ (no integer cube of 6)

• For $n = 3$: $n^2 = 9$ $m^3 = 11$ (no integer cube of 11)

• For $n = 4$: $n^2 = 16$ $m^3 = 18$ (no integer cube of 18)

• For $n = 5$: $n^2 = 25$ $m^3 = 27$ (integer cube of 27, i.e., $m = 3$)

Thus, for $n = 5$, we have $n^2 = 25$ and $m^3 = 27$. Therefore, Mr. X's age last year was 25, and his current age is: $x = 25 + 1 = 26$

Step 2: Find when his age becomes the cube of a number again

We need to find when his age becomes the cube of a number again. That is, when $x + t = k^3$ for some integer $k$, where $t$ is the number of years Mr. X has to wait.

Since his current age is 26, we need to find the smallest $t$ such that: $26 + t = k^3$

Let's try different values of $k$:

• For $k = 3$: $k^3 = 27$ $t = 27 - 26 = 1$ (but 27 is next year, not again in future)

• For $k = 4$: $k^3 = 64$ $t = 64 - 26 = 38$

Thus, the least number of years Mr. X must wait for his age to become the cube of a number again is 38.

Answer: B. 38