The maximum segment lifetime (MSL) is the maximum amount of time a TCP segment can exist in the network before it is considered expired and discarded. In the case of this question, the MSL is given as 60 seconds.

To calculate the minimum number of bits required for the sequence number field in a TCP connection, we need to consider the following factors:

1. Bandwidth: The bandwidth of the connection is given as 1 Gbps (Gigabits per second).
2. MSL: The MSL is given as 60 seconds.

Now, let's break down the calculation step-by-step:

1. Calculate the maximum number of bits that can be transmitted in 60 seconds:
- Bandwidth is given in Gigabits per second (Gbps), so we need to convert it to bits per second (bps).
- 1 Gbps = 1,000,000,000 bps.
- Multiply the bandwidth by the MSL to get the maximum number of bits transmitted in 60 seconds:
1,000,000,000 bps * 60 seconds = 60,000,000,000 bits.

2. Calculate the range of sequence numbers:
- The sequence number field is used to uniquely identify each TCP segment.
- The range of sequence numbers is determined by the number of bits available for the sequence number field.
- The range is calculated as 2^n, where n is the number of bits.
- We need to find the value of n that satisfies the condition 2^n >= 60,000,000,000.

3. Find the minimum number of bits required:
- Start with n = 1 and increment it until we find a value that satisfies the condition.
- By trying different values, we find that 2^33 = 8,589,934,592, which is greater than 60,000,000,000.
- Therefore, the minimum number of bits required for the sequence number field is 33.

In conclusion, the minimum number of bits required for the sequence number field in a TCP connection to send a maximum segment lifetime of 60 seconds with a bandwidth of 1 Gbps (without wrap around time) is 33 bits.