Answer: B — $9$

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

eq b$.

Since $B$ is obtained by reversing the digits of $A$, we have:

$B=10b+a$

Since $B$ is also a two-digit number, $b

eq 0$.

Now,

$A-B=(10a+b)-(10b+a)=9a-9b=9(a-b)$

It is given that $A-B$ is a multiple of $27$. Therefore:

$9(a-b) \text{ is a multiple of } 27$

So, $a-b$ must be a positive multiple of $3$.

Since $A>B$, we have $a>b$. The possible values of $a-b$ are:

$3, 6, 9$

Now count the valid digit pairs with $a>b$ and $b

eq 0$:

For $a-b=3$:

$(a,b)=(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)$

This gives $6$ numbers.

For $a-b=6$:

$(a,b)=(7,1),(8,2),(9,3)$

This gives $3$ numbers.

For $a-b=9$, the only possible pair is $(9,0)$, but this is not allowed because $B$ would not be a two-digit number.

Hence, the total number of possible values of $A$ is:

$6+3=9$

Therefore, the correct answer is $9$.