Let $S = \{1, 2, 3, \dots, 60\}$. Relation $R = \{(a,b) : b=pq, p,q \geq 3, p,q \text{ are primes}\}$.
Prime numbers greater than or equal to $3$ and less than or equal to $60$ are $3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59$.

We need to find values of $b = p \times q \leq 60$.

If $p=3$, then $q$ can be $3, 5, 7, 11, 13, 17, 19$. That gives us $7$ possibilities.
If $p=5$, then $q$ can be $5, 7, 11$. That gives us $3$ possibilities.
If $p=7$, then $q$ can be $7$. That gives us $1$ possibility.

Total possibilities are $7+3+1 = 11$.

Since $a$ can be any number in the set $S$, there are $60$ possible values for $a$ for each of the $11$ values of $b$.
Therefore, total number of elements in the relation $R$ is $60 \times 11 = 660$.