The age of Mr. X last year was the square of a number, and it would be the cube of a number next year. What is the least number of years he must wait for his age to become the cube of a number again?

The correct option is 38.

Let us analyze the details given in the question step-by-step to find the present age of Mr. X and determine how many years he must wait for his age to become a cube of a number again.

Step 1: Determine Mr. X's current age.
Let Mr. X's current age be represented as $A$.

According to the problem statement:
1. His age last year was a perfect square: $A-1=x^2$
2. His age next year will be a perfect cube: $A+1=y^3$

From these two equations, the difference between the next year's age and last year's age is:
$(A+1)-(A-1)=y^3-x^2$
$2=y^3-x^2$
$y^3-x^2=2$

We need to find a perfect cube ($y^3$) and a perfect square ($x^2$) that differ by 2.
Let us test small values for perfect cubes:
• If $y=1$, $y^3=1$. Then $x^2=1-2=-1$ (not possible).
• If $y=2$, $y^3=8$. Then $x^2=8-2=6$ (6 is not a perfect square).
• If $y=3$, $y^3=27$. Then $x^2=27-2=25$ (25 is a perfect square, since $5^2=25$).

Thus, $x=5$ and $y=3$ satisfies the conditions:
• Age last year: $25=5^2$
• Age next year: $27=3^3$
• Current age ($A$): $26$ years.

Step 2: Calculate the waiting time for his age to become a cube again.
The current cube age next year is $3^3=27$ years.
The very next perfect cube after $3^3$ is $4^3$:
$4^3=64$

To find the least number of years he must wait from his present age of 26 years to reach 64 years:
$\text{Waiting years}=64-26=38$ years.

Therefore, Mr. X must wait 38 years for his age to become a cube of a number again.