Let the two-digit number $A$ be represented as $10a + b$, where $a$ and $b$ are the digits of $A$ and $a

eq b$. The number$B$, obtained by reversing the digits of$A$, is$10b + a$.

Given that $A > B$, we have: $A - B = (10a + b) - (10b + a) = 9a - 9b = 9(a - b).$

We are told that $A - B$ is a multiple of $27$. Therefore, $9(a - b)$ must be a multiple of $27$.

This implies: $a - b = \frac{27}{9} = 3.$

Thus, $a - b = 3$. Since $a$ and $b$ are digits (0 through 9), we need to find pairs $(a, b)$ such that $a - b = 3$ and $a > b$.

The possible pairs $(a, b)$ are:

• $(3, 0)$: $A = 30$
• $(4, 1)$: $A = 41$
• $(5, 2)$: $A = 52$
• $(6, 3)$: $A = 63$
• $(7, 4)$: $A = 74$
• $(8, 5)$: $A = 85$
• $(9, 6)$: $A = 96$

Thus, there are 6 such numbers $A$ that satisfy the condition.

Therefore, the correct option is A.