There are two questions here:
How many triples of numbers $(X,Y,Z)$ add to 20? We can call this $A$.
How many of these are divisible by 5? We can call this $B$.
The answer will then be $B/A$.
First we calculate $A$. Let us write $C(x)$ for the total number of ways of choosing $(X,Y,Z)$ so that $X=x$ and $X+Y+Z=20$.
There are 21 possible choices for $X$. For each such $X$, there are $21-X$ possible choices for $Y$ that make $X+Y\le 20$, namely $\{0, 1, \ldots 20-X\}$. The other choices have $X+Y>20$ and therefore $X+Y+Z>20$. Once we have chosen $X$ and $Y$ at random, there is exactly one possible choice for $Z$ that makes $X+Y+Z=20$. So we have $C(x) = 21-x$.
We want the total of $C(x)$ for each $x$ between 0 and 20: $\begin{eqnarray} &&\sum_{x=0}^{20} C(x) \\ &=& \sum_{x=0}^{20} (21-x)\\ &=& \sum_{x=0}^{20} 21 - \sum_{x=0}^{20} x \\ &=& 441 - 210 \\ &=& 231. \end{eqnarray}$
Now we calculate $B$. 5 is prime, so $XYZ$ is divisible by 5 if and only if one of $X$, $Y$, or $Z$ is divisible by 5. We can use inclusion-exclusion: $B$ is the sum of the cases where (at least) $X$, $Y$, or $Z$ is divisible by 5, minus the cases where (at least) two are divisible by 5, plus the cases where all three are divisible by 5. That is, $\begin{eqnarray}B&=&D_x + D_{y} + D_z \\ &&- D_{xy} - D_{xz} - D_{yz} \\ &&+ D_{xyz}\end{eqnarray}$
Where $D_{xy}$ denotes the number of choices of $(X,Y,Z)$ where $5\mid X$ and $5\mid Y$, and similarly for the others.
By symmetry, $D_x = D_y = D_z$, and $D_{xy} = D_{xz} = D_{yz}$. Also, it is impossible to have two of $(X,Y,Z)$ divisible by 5 without the third also being divisible by 5, so $D_{xy} = D_{xyz}$. So the previous equation reduces to:
$B = 3D_x - 2D_{xyz}$
We calculate $D_x$: $X$ will be a multiple of 5 whenever $X\in\{0,5,10,15,20\}$, so we want $\begin{eqnarray} &&\sum_{5\mid X} C(X) \\ &=& \sum_{X\in\{0,5,10,15,20\}} (21-X) \\ &=& 21\cdot5 - (0+5+10+15+20) \\ &=& 105 - 50 \\ &=& 55 \end{eqnarray}$
To calculate $D_{xyz}$ is quick because there are very few such triples, and we can enumerate them by brute force: $(0,0,20)\ldots (0,20,0), (5,0,15)\ldots (5,15,0),\ldots (20,0,0)$. This is 5+4+3+2+1 = 15.
So we have $B = 3\cdot 55 - 2\cdot 15 = 135$.
Thus the answer is $B/A = 135/231 = 45/77$.