Let A = Event that exactly one odd number turns up
So,
$P(A) = \frac{n(A)}{n(S)} $
We have rolled a die 3 times. Assuming the rolls are independent, the outcomes will be in the form
( _ )( _ )( _ )
So, for the sample space there are $(6).(6).(6) = 6^3$ outcomes.
Now, when we roll a die, the outcomes are $\{ 1,3,5\}$ and $\{2, 4, 6 \}$
For event A, we want exactly one odd number. So, the resultant outcomes will be in the form
[odd](even)(even) OR (even)[odd](even) OR (even)(even)[odd]
So, for each of the above there are 3 ways to choose a odd number from $\{ 1, 3,5\}$ AND 3 ways to choose an even number from $\{2,4,6\}$ AND 3 ways to choose an even number from $\{2,4,6\}$, and hence the total number of outcomes for event A will be
$[3]\cdot (3) \cdot (3) + (3) \cdot [3] \cdot (3) + (3) \cdot (3) \cdot [3] = 3 \cdot 3^3$
because, AND means multiplication and OR means addition
So, $P(A) = \frac{3\cdot 3^3}{6^3} = 3 \cdot (\frac{3}{6})^3 = 3 \cdot (\frac{1}{2})^3 = \frac{3}{8}$