There are indeed $\binom{52}{13}$ different 13-card hands and this will indeed be the size of our sample space and thus our denominator when we finish our calculations.
For the numerator, we need to pause for a moment and understand what the problem is actually asking, since this appears to be where you got stuck.
We are asked to find the probability that in our hand of thirteen cards, there is at least one suit for which we have all three face cards. For example $(A\spadesuit,2\spadesuit,3\spadesuit,\dots,10\spadesuit,J\spadesuit,Q\spadesuit,K\spadesuit)$ has all three of the face cards for spades. Similarly if all those cards happened to be hearts instead it would also count since we would have all of the face cards for hearts. Similarly still, a hand like $(J\spadesuit,Q\spadesuit,K\spadesuit,J\heartsuit,Q\heartsuit,K\heartsuit,J\diamondsuit,Q\diamondsuit,K\diamondsuit,\dots)$ would count since we have all of the face cards from spades (we also happen to have all of the face cards from hearts and diamonds too).
Let $\spadesuit$ represent the event that we have have all of the face cards from spades. Similarly, let $\diamondsuit, \heartsuit, \clubsuit$ represent the event that we have all of the face cards from diamonds, hearts, and clubs respectively.
You are asked to find $Pr(\spadesuit\cup\diamondsuit\cup \clubsuit\cup \heartsuit)$
To do this, let us apply inclusion exclusion. We expand the above as:
$Pr(\spadesuit\cup \diamondsuit\cup\clubsuit\cup\heartsuit) = Pr(\spadesuit)+Pr(\diamondsuit)+\dots-Pr(\spadesuit\cap \diamondsuit)-Pr(\spadesuit\cap \clubsuit)-\dots+Pr(\spadesuit\cap \diamondsuit\cap \clubsuit)+\dots-Pr(\spadesuit\cap\diamondsuit\cap \clubsuit\cap \heartsuit)$
Now, let us calculate each individual term in the expansion.
The calculation you did before is relevant. Indeed, we calculate $Pr(\spadesuit)=\dfrac{\binom{3}{3}\binom{49}{10}}{\binom{52}{13}}$. This is again merely the probability that we have all of the face cards from the spades and is not the final probability that we were tasked with calculating.
We continue and calculate more terms:
For example $Pr(\spadesuit\cap \diamondsuit)=\dfrac{\binom{6}{6}\binom{46}{7}}{\binom{52}{13}}$
We then notice what symmetry there is in the terms and can simplify some. Finally, we write the final expression for our final answer (and get an exact number only if actually requested or required, usually opting to leave the answer in terms of binomial coefficients without additional simplification).
Best Answer
What is your source? The expression $\frac{4}{\binom{52}{5}}$ is correct (you don't need $\binom{4}{1}$ at the top; one easily counts the hands directly with no calculation!).
I suspect your source is wrong. E.g.
https://books.google.com/books?id=opjqt3SbE9MC&pg=PA81&dq=%22royal+flush%22+odds&hl=en&sa=X&ved=0ahUKEwjF9vqXzIbTAhWh1IMKHXSGA3EQ6AEIIDAB#v=onepage&q=%22royal%20flush%22%20odds&f=false
gives "odds of getting any one of these royal flushes are 2,598,956 to 4" and since $\binom{52}{5} = 2,598,960$, but "odds" are given as fail to successes or successes to fails, so this is a probability of $\frac{4}{2598956 + 4} = \frac{4}{2598960} = \frac{4}{\binom{52}{5}}$.
Other sources give the same result, and it's the only one that makes mathematical sense.