In problems like this we always choose the most restrictive things first:

The most restrictive thing is the third digit.  Since the number must be odd, we
must choose the third digit odd.  There are only 3 ways to choose the third
digit odd, 1,3, or 5

That's 3 ways to choose the third digit.

Now let's choose the first digit:

Since we can't repeat the digit we used for the third digit, then for each of
those 3 ways to choose the third digit, there are only 4 ways left to choose the
first digit.

Therefore the number of ways to choose the third and first digits is 3×4 or 12
ways.

Now let's choose the second digit:

Since we can't repeat either of the two digits we used for the third and first
digits, then for each of those 12 ways we could choose the third and first
digits there are only 3 ways left to choose the second digit. 

Therefore the number of ways to choose the three digits is 3×4×3 = 36.

Here's a computer check.  6 rows of 6 each makes 36.

123 125 135 143 145 153 
213 215 231 235 241 243 
245 251 253 315 321 325 
341 345 351 413 415 421 
423 425 431 435 451 453 
513 521 523 531 541 543 

Edwin