1. Understand the relation

We have a relation $R$ from the set $A=\{1,2,3,\ldots,60\}$ to itself, defined by $R=\{(a,b): b=pq,\text{ where } p,q\ge 3 \text{ are primes}\}.$

Here, $a$ can be any element of $A$, while $b$ must be an element of $A$ that can be written as a product of two primes, each at least $3$.

So the total number of ordered pairs in $R$ is: $|R|=(\text{number of choices for }a)\times(\text{number of valid }b).$

Since $a\in A$, there are clearly $60$ choices for $a$.

Thus we only need to count how many numbers $b\in \{1,2,\dots,60\}$ can be written as $b=pq$ with $p,q\ge 3$ prime.

2. List primes at least 3 up to 60

The relevant primes are: $3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59.$

We need products $pq\le 60$.

3. Find all possible products $pq\le 60$

We allow $p=q$ as well.

• With $p=3$: \quad 3\cdot 5=15, \quad 3\cdot 7=21, \quad 3\cdot 11=33, \quad 3\cdot 13=39, \quad 3\cdot 17=51, \quad 3\cdot 19=57.$$ Next, $3\cdot 23=69>60$, so stop.
• With $p=5$: \quad 5\cdot 7=35, \quad 5\cdot 11=55.$$ Next, $5\cdot 13=65>60$, so stop.
• With $p=7$: $7\cdot 7=49.$ Next, $7\cdot 11=77>60$, so stop.

For larger primes, even the square exceeds $60$, so no more values occur.

Hence the valid values of $b$ are: $\{9,15,21,25,33,35,39,49,51,55,57\}.$

So the number of valid $b$ is $11.$

4. Count ordered pairs

For each of the $11$ valid values of $b$, $a$ can be any of the $60$ elements.

Therefore, $|R|=60\times 11=660.$

5. Check options

The correct option is $\boxed{660}$ which is Option B.

6. Compare with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.