Here is a statement of the question, which is taken from Chaisson and McMillan's Astronomy: A Beginner's Guide to the Universe.
Assume that the number of (technological) civilisations in the Milky Way Galaxy is equal to the average lifetime of a civilisation. Assuming that civilisations are uniformly spread over a two-dimensional galactic disk of radius 15 kiloparsecs, and all have the same lifetime, calculate the minimum lifetime for which a two-way radio communication with our nearest neighbour would be possible before our civilisation ends. Repeat the calculation for a round-trip personal visit, using a spacecraft that travels at 50 kilometres per second.
Here is my approach to the problem:
Let $n$ be the number of civilisations in the Milky Way and let $k$ be the average lifetime of a civilisation in years. By hypothesis, $n=k$. We will also assume that the galaxy is overlaid with a grid of squares, each of which has a single civilisation at its centre.
The area of the galactic disk is $15^2 \pi =225\pi $ kiloparsecs. Since the civilisations are distributed uniformly, it follows that each civilisation is centred in a square whose area is $\frac{225\pi }{n} $ kiloparsecs. If $s$ is the length of a side of such a square in kiloparsecs, then $s$ is also the distance between the centres of two adjacent squares, hence, two civilisations. From the structure of the grid, $s$ represents the shortest distance between any pair of civilisations in the Milky Way. Note that $s^2= \frac{225\pi }{n} $ and hence $s=\frac{15\sqrt{\pi }}{\sqrt{n} }$ kiloparsecs.
The minimum roundtrip distance between any pair of civilisations is therefore $2s=\frac{30\sqrt{\pi }}{\sqrt{n}} $ kiloparsecs. If $v$ is the velocity in kilometres per second taken for a roundtrip between two such civilisations, then the time, $t$ taken in seconds to make that roundtrip is given by $$t=\frac{2s\cdot 3.09 \cdot 10^{16} }{v} = \frac{30\sqrt{\pi } \cdot 3.09 \cdot 10^{16} }{v\sqrt{n}}.$$
We require that $$t\leq 365\cdot 24\cdot 60^2 k .$$ Recalling that $k=n$, we compute: \begin{equation} \begin{split} \frac{30\sqrt{\pi } \cdot 3.09 \cdot 10^{16} }{v\sqrt{k}} & \leq 365\cdot 24\cdot 60^2 k \\ \frac{30\sqrt{\pi } \cdot 3.09 \cdot 10^{16} }{365\cdot 24\cdot 60^2 v } & \leq k^\frac{3}{2} \\ \Big{(}\frac{30\sqrt{\pi } \cdot 3.09 \cdot 10^{16} }{365\cdot 24\cdot 60^2 v }\Big{)}^\frac{2}{3} & \leq k . \end{split} \end{equation}
For radio communication, $v= c=3\cdot 10^5 $ kilometres per second. Inserting this into the inequality, we see that correct to two significant figures $k\geq 3100 $ years.
For our spacecraft, $v=50$ kilometres per second. Inserting this into the inequality, we see that $k\geq 1000000 $ years.
I chose to use a square grid since my answer for radio communication agreed with Chaisson and McMillian's answer when I used that approach. I think one gets similar numbers when one overlays the galaxy with equilateral triangles or regular hexagons, but squares seemed the simplest. Using circles also gave a similar result, but it is of course harder to cover the whole galaxy that way. Unfortunately, Chaisson and McMillian gave an answer of 3000000 years for the spacecraft situation. Therefore, either my approach is seriously flawed, or there is a typo in the answers. Both are entirely possible. Can anyway spot a serious flaw in my approach?