Solve the Fraction Equation: Finding the Numerator in ?/5a = 2/5b

$$Let's examine the problem:

$\frac{?}{5a}=\frac{2}{5b}$

Now let's think logically, and remember the fraction reduction operation,

For the fraction on the left side to be reducible, we want all the terms in its numerator to have a common factor. Additionally, note that the number 5 (which is in the denominator of the fraction on the right side) already exists in the denominator of the fraction on the left side, so we don't want to reduce it,

We'll just add that we want to get the number 2 that appears in the numerator of the fraction on the right side,

Now, we want to reduce the term $a$ from the denominator of the fraction on the left side since it does not appear in the denominator on the right side and simultaneously get the term $b$ in the denominator of the fraction on the right side, note that this term does not appear in the expression in the denominator of the fraction on the left side, therefore we will choose the expression:

$\frac{ 2a}{b}$

Let's verify that from this choice we will indeed get the expression on the right side, we'll use the fact that multiplying a number by a fraction is actually multiplying the number by the fraction's numerator (in the first stage), and in fraction multiplication (in the second stage) in order to simplify the fraction resulting from this choice, then we'll reduce the simplified fraction:

$\frac{?}{5a}=\frac{2}{5b} \\ \downarrow\\ \frac{\big(\frac{2a}{b}\big)}{5a}\stackrel{?}{= }\frac{2}{5b} \\ \frac{1}{5a}\cdot\frac{2a}{b}\stackrel{?}{= }\frac{2}{5b}\\ \frac{2\textcolor{red}{\not{a}}}{5\textcolor{red}{\not{a}}b}\stackrel{?}{= }\frac{2}{5b} \\ \downarrow\\ \boxed{\frac{2}{5b} \stackrel{!}{= }\frac{2}{5b} }$

Therefore this choice is indeed correct.

In other words - the correct answer is answer D.