1. We need to form three-digit even numbers using the digits $\{0,1,3,4,6,7\}$ without repetition.

2. For a number to be even, its unit digit must be one of: $0,4,6$

3. We count case by case.

Case 1: Unit digit = 0

• Fix the unit digit as $0$.
• The hundreds digit can be chosen from $\{1,3,4,6,7\}$, so there are $5$ choices.
• After choosing the hundreds digit, the tens digit can be chosen from the remaining $4$ digits.

So, number of such numbers: $5 \times 4 = 20$

Case 2: Unit digit = 4

• Fix the unit digit as $4$.
• The hundreds digit cannot be $0$, and cannot be $4$.
• So the hundreds digit can be chosen from $\{1,3,6,7\}$, giving $4$ choices.
• The tens digit can then be chosen from the remaining $4$ digits (including $0$ if unused).

So, number of such numbers: $4 \times 4 = 16$

Case 3: Unit digit = 6

• Fix the unit digit as $6$.
• The hundreds digit cannot be $0$, and cannot be $6$.
• So the hundreds digit can be chosen from $\{1,3,4,7\}$, giving $4$ choices.
• The tens digit can then be chosen from the remaining $4$ digits.

So, number of such numbers: $4 \times 4 = 16$

4. Total number of three-digit even numbers: $20 + 16 + 16 = 52$

Therefore, the required number is $\boxed{52}$