Resistances n, each of r ohm, when connected in parallel give an equivalent resistance of R ohm. If these resistances were connected in series, the combination would have a resistance in ohms, equal to

For n equal resistances r in parallel: $\dfrac{1}{R} = \dfrac{1}{r} + \dfrac{1}{r} + \dots$ (n times) $= \dfrac{n}{r}$
So according to the question $R = \dfrac{r}{n} \Rightarrow r = nR$
For the same n resistances in series: $R_{series} = r + r + \dots$ (n times) $= nr$
Substituting $r = nR$: $R_{series} = n \times nR = n^2R$
So the series combination has a resistance of $n^2R$ ohm.