How many three digit numbers exist such that the third digit is the geo mean How many three digit numbers exist such that one of the digits is the geometric mean of the other two?
  A 12, B 18, C 24, D other


So, $N = 100a + 10b + c$ let $c =\sqrt{ab}$. 

$ab$ must be a perfect square, so $ab = 1, 4, 9, 16, 25, 36, 49, 64, 81$

Case: $ab = 1 \implies a=1, b=1$. 

Case: $ab = 4 \implies a = 2, b=2$

Case: $ab = 9 \implies a=3, b=3$.

Case: $ab = 16 \implies (a=8, b=2), (a=4, b=4)$

Case: $ab = 25 \implies a=5, b=5$

Case: $ab = 36 \implies (a=9, b=4), (a=6, b=6)$

Case: $ab = 49 \implies (a=7, b=7)$

Case: $ab = 64 \implies (a=8, b=8)$

Case: $ab=81 \implies (a=9, b=9)$.

$(a=8, b=2, c = 4)$ also means that we could replace this with, $ (a, b, c) = (2, 4, 8) = (4, 2, 8) = (4, 8, 2)$

$(a=9, b=4, c=6)$ also means $(a, b, c) = (9, 4, 6) = (4, 6, 9) = (9, 6, 4) = (4, 9, 6) = (6, 9, 4) = (6, 4, 9)$.

So to me, the answer is B $18$
You asked two different questions.  The question in the title appears to be: 


  "How many three digit numbers exist such that the third digit is the
  geometric mean of the other two?"


I will answer the question in the statement of the problem: 


  "How many three digit numbers exist such that one of the digits is the
  geometric mean of the other two?"


Let $x, y \geq 0$.  The geometric mean of $x$ and $y$ is $\sqrt{xy}$.  

A three digit positive integer is a number of the form $100a + 10b + c$, where $a \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ and $b, c \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.  

We consider cases:


The geometric mean of the digits is $0 \implies 0 = \sqrt{a \cdot 0}$, where $a \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.  There are nine possibilities: $100, 200, 300, 400, 500, 600, 700, 800, 900$.
If $n \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $n = \sqrt{n \cdot n}$.  There are nine such numbers.  They are $111, 222, 333, 444, 555, 666, 777, 888, 999$.
$2 = \sqrt{1 \cdot 4}$.  Then the number is one of the $3! = 6$ permutations of the digits $1$, $2$, and $4$.  They are $124, 142, 214, 241, 412, 421$.
$3 = \sqrt{1 \cdot 9}$.  Then the number is one of the $3! = 6$ permutations of the digits $1$, $3$, and $9$.  They are $139, 193, 319, 391, 913, 931$.
$4 = \sqrt{2 \cdot 8}$.  The number is one of the $3! = 6$ permutations of the digits $2$, $4$, and $8$.  They are $248, 284, 428, 482, 824, 842$.
$6 = \sqrt{4 \cdot 9}$.  The number is one of the $3! = 6$ permutations of the digits $4$, $6$ and $9$.  They are $469, 496, 649, 694, 946, 964$.  


Hence, there are $2 \cdot 9 + 4 \cdot 6 = 42$ three digit numbers such that one of the digits is the geometric mean of the other two.
Third digit is $\sqrt{ab}$ implies square root should yield single digit so it can be $[0,9]$ thus now digits whose product is 0 so $100,..900$ , for 1 is only $1,1$ so number $111$ then for 2 so product is $4$ thus numbers are $222,142,412$ then $3$ so $333,193,913$ now $4$ so $284,824,444$ then $5$ so $555$ then $6$ so $496,946,666,$  for $7$ so $777$ then $8$ so $888$ and finally $9$ so $999$ thus total numbers are $9+1+3+3+3+1+3+1+1+1=26$ so its $D$