The parking problem riddle

According to http://mathworld.wolfram.com/RenyisParkingConstants.html, if $M(n)$ is the expected number of cars of length 1 to fit onto a stripe of length $n$, you have $$ m := \lim_{n\to\infty} \frac{M(n)}{n} \approx 0.7476 \text{.} $$

In other words, the expected density (i.e., the total length of the parked cars divided by the length of the pavement) goes to $0.7476\ldots$ as the pavement length goes to infinity. In your case that leads to an estimate for the expected number $N$ of 3m long cars on a 300m long pavement of $$ \frac{3N}{300} \approx 0.7476 \implies N \approx 74.76 \text{.} $$

A more precise estimate from that page is $$ M(n) \approx mn + m - 1 \text{,} $$ again for the expected number of 1m long cars on a stripe of length $n$. To apply that to your question, one has to use that the expected number $N$ of 3m long cars on a 300m long stripe is the same as the expected number of 1m long cars on a 100m long stripe, which yields $$ N \approx M(100) \approx 74.51 \text{.} $$

Let $f(r)$ be the expected number of cars that will park in an interval of length $r$. By considering what happens when the first car comes along, we should have $$ f(r) = \cases{ 0 & if $r < 1$\cr 1 + \frac{1}{r-1} \int_0^{r-1} ds\; (f(s) + f(r-1-s))\cr \ = 1 + \frac{2}{r-1} \int_0^{r-1} ds \; f(s) & otherwise}$$ So:

If $1 \le r < 2$, $f(r) = 1$.

If $2 \le r < 3$, $f(r) = 1 + \frac{2 (r-2)}{r-1} = 3 - \frac{2}{r-1} $.

etc.

Unfortunately, by the time we get to $6 < r < 7$, the integrals can't be done in closed form.