Interpreting the ratio of two variances in a Bayesian decision problem
Excuse the word-play, but the interpretation of $n'$ is a... posterior one. Meaning, the important thing is not how $n'$ is defined (ratio of variances, although this will prove consistent with the interpretation), but how it functions in the posterior mean and variance.  

What does it do? For the posterior variance, it is easiest: firstly, it appears as an addition to the expression's denominator, an addition comparable to the actual sample size. This permits the authors to talk about "equivalent number of sample observations". Secondly, its effect is to lower the variance, which is a metric of dispersion but also of uncertainty. This effect now permits the authors to talk about "the amount of information implicit in the prior distribution": more information in the prior - lower uncertainty in the prior - lower uncertainty in the posterior - and indeed, $\sigma^2_0 \downarrow \implies n' \uparrow \implies \text{Posterior Variance} \downarrow$.  

For the posterior mean we have (setting $N \equiv  n' + n$)

$$\text {Posterior Mean} = \frac{n'\mu_0 +\sum_{i=1}^nx_i}{n'+n} = 
\frac{n'}{N}\mu_0 + \frac{n}{N}\bar x$$

i.e. a convex combination of the prior mean and the sample mean - so the magnitude $n'$ weighs the prior information against the sample information, as though you had two separate samples from the same population of sizes $n'$ and $n$, and considered their pooled average. Again, this can be interpreted as "information implicit in the prior distribution expressed in terms of virtual/fictitious sample size".