To solve this problem, we start by using the information given regarding the relationship between rate, time, and distance.
Since rate varies inversely with time, we can express this relationship as follows:
\[ \text{Rate} \times \text{Time} = \text{Constant} \]
Let \( d \) be the distance traveled. From the problem, we know that a car travels this distance at a rate of 165 mph in \( \frac{3}{4} \) minute.
First, we convert the time from minutes to hours because the speed is given in miles per hour (mph):
\[ \frac{3}{4} \text{ minute} = \frac{3}{4} \times \frac{1}{60} \text{ hours} = \frac{3}{240} \text{ hours} = \frac{1}{80} \text{ hours} \]
Next, we can find the distance \( d \) traveled by using the formula:
\[ d = \text{rate} \times \text{time} = 165 \text{ mph} \times \frac{1}{80} \text{ hours} \]
Calculating this gives:
\[ d = 165 \times \frac{1}{80} = \frac{165}{80} = \frac{33}{16} \text{ miles} \]
Now, we need to find the rate required to travel the same distance \( d \) in \( \frac{1}{3} \) minute. We'll first convert \( \frac{1}{3} \) minute to hours:
\[ \frac{1}{3} \text{ minute} = \frac{1}{3} \times \frac{1}{60} \text{ hours} = \frac{1}{180} \text{ hours} \]
Let \( r \) be the rate needed to travel the same distance in this new time. Using the relationship \( \text{Rate} \times \text{Time} = d \), we have:
\[ r \times \frac{1}{180} = \frac{33}{16} \]
Now, solving for \( r \):
\[ r = \frac{33}{16} \times 180 \]
Calculating this gives:
\[ r = \frac{33 \times 180}{16} = \frac{5940}{16} = 371.25 \text{ mph} \]
Therefore, the rate needed to go the same distance in \( \frac{1}{3} \) minute is \( \boxed{371.25} \) mph.