In 2018

The nearest perfect square number greater than or equal to $2108$ is $2025$ . $\implies 45^2 = 2025$

Now, if the age in year $2025$ is $45$ then the age in year $2018$ is :

$44 - (2025 - 2018) + 1$

$\implies 44 - 7 + 1 = \color{#20A900}38$

In order to be $x$ years old in the year $x^2$ , one would have to be born in the year $x^2 - x = x(x - 1)$ . Also, $x$ should be close to $\sqrt{2018} \approx 44.92$ .

• $x = 44$ : To be $44$ in the year $44^2 = 1936$ , one would have to be born in the year $(44)(43) = 1892$ , making the person $126$ years old in $2018$ ; likely impossible (*) .
• $x = 45$ : To be $45$ in the year $45^2 = 2025$ , one would have to be born in the year $(45)(44) = 1980$ , making the person $38$ years old in $2018$ ; possible .
• $x = 46$ : To be $46$ in the year $46^2 = 2116$ , one would have to be born in the year $(46)(45) = 2070$ , so they wouldn't be born yet in $2018$ ; impossible .

So the only reasonable answer is $x = \boxed{45}$ .

(*) : A quick online search shows the oldest verified human age to be 122, though there are claims as high as 160.