There are a couple of ways to do this question. One way is to bit quite `pure' about it and start trying to bound the series. You might, for example, bound $|sin (2n+1)\pi x| \leq 1$, and then try to somehow make a comparison to a series for which you

*do* know the sum. For example, you may want to compare it to a series
\[
\sum_{n=0}^\infty r^n = \frac{1}{1-r},
\]

with a specific choice of $r < 1$, and then see if you can use that bound to select the $t$. Perhaps a less technical approach is to first gather some intuition. First, you realize through the physics or through the formula that the maximum of the temperature must be at $x = 1/2$. Therefore, we only need to examine
\[
T(\tfrac{1}{2},t) = \sum_{n=0}^\infty \left[\frac{4}{\pi(2n + 1)}\right] \sin (2n+1)\frac{\pi}{2} e^{-(2n+1)^2\pi^2 t}.
\]

Next, examine the below figure, which plots the above equation for $n = 0$ and $n = 10$.

We see that it looks like the time where the solution first dips is somewhere around $t \approx 0.1$. In fact, it almost looks like we can just use *a single term* from the approximation to get an accurate answer. That is, can we try:
\[
\frac{4}{\pi} e^{-\pi^2 t } = 0.5 \Rightarrow t = -\frac{1}{\pi^2} \log(0.5 \pi/4) \approx 0.0947061.
\]

With a $n = 100$, we get $t^* \approx 0.0947061$, so in fact our approximation using $n = 0$ has a relative error of only $2\%$(!) Why did it work out so well? It worked well because as soon as the argument of the exponential is $< 1$, the decay of the later terms of the approximation is dominant. The most rough approximation is to simply require that
\[
\frac{1}{\pi^2 (2n+1)^2}\rvert_{n = 0} t = 1,
\]

which gives
\[
t = \frac{1}{\pi^2} \approx 0.101321,
\]

which gets us in the appropriate ballpark. The little adjustment then comes from the first pre-factor of the series expansion.

Finishing the question, in dimensional terms, this would correspond to $t^* = 0.095 L^2/\kappa$. If we expand the domain and send $L = 2\ell$, this would multiply the time by a factor of $4$. *Cooking tip: so if you double the size of a long cake, you have to cook it for four times as long?*