The probability that a specific bulb dies in time $\le 3$ is $1-e^{-3/800}$. Call this $p$.
Then the probability that the fourth shortest-lived bulb has life $\le 3$ is $1$ minus the probability that $3$ or fewer bulbs are dead by that time.
This can be calculated by the standard formula for the binomial distribution.
We can alternately use the Poisson approximation to the binomial, or by direct use of the Poisson. The mean number of "deaths" in $3$ hours is $1000\cdot\frac{3}{800}$. Let $\mu$ be this number. Then the probability of $k$ deaths in $3$ hours is approximately $e^{-\mu}\frac{\mu^k}{k!}$. Calculate the sum of the probabilities, $0$ to $3$, and subtract the sum from $1$.
Remark: Lifetimes of the bulbs, at least if they are installed in the same building, are not independent. Think power surge!
Also, the exponential is not a really good model for bulb lifetime. There is a peak of early deaths, because of manufacturing defects. And the exponential distribution is appropriate for things that die but do not age, like radioisotopes. Lightbulbs age.
If there were just one light bulb with an average lifetime of $1000$ hours, then the number of failures in time $t$ would have a Poisson distribution with expected value $t/1000$. With $10$ light bulbs functioning independently of each other, that expected value is simply multiplied by $10$. The reason is that the distribution of the sum of independent Poisson-distributed random variables is Poisson-distributed. I think the question of why that is so has been answered here a number of times.
Best Answer
Let random variable $X$ be the lifetime of the bulb.
The density function of $X$ is $\frac{1}{1000}e^{-x/1000}$ (for $x\ge 0$). We want to find $m$ so that $$\Pr(X\gt m)=\frac{1}{2}.$$ So we want $$\int_m^\infty \frac{1}{1000}e^{-x/1000}\,dx=\frac{1}{2}.$$ Integrate. We get $$e^{-m/1000}=\frac{1}{2}.$$ Take the natural logarithm of both sides. We get $$-\frac{m}{1000}=\ln(1/2)=-\ln 2.$$ It follows that $m=1000\ln 2$.
Remark: The exponential distribution fits the lifetime of real-life incandescent bulbs quite poorly.