Answer:
Approximately [tex]1.04\; \rm L[/tex] of the [tex]12.0\; \rm M[/tex] [tex]\rm NaOH[/tex] solution is needed. Approximately another [tex]3.96\; \rm L[/tex] of water will also be required.
Assumption: the volume of these [tex]\rm NaOH[/tex] solutions does not depend on the quantity of [tex]\rm NaOH\![/tex] (the solute) that each of them contain.
Explanation:
Calculate the number of moles of [tex]\rm NaOH[/tex] formula units in that [tex]5.00\; \rm L[/tex] of [tex]2.50\; \rm M[/tex] [tex]\rm NaOH\![/tex] solution:
[tex]n(\mathrm{NaOH}) = c\cdot V = 5.00\; \rm L \times 2.50\; \rm mol \cdot L^{-1} = 12.5\; \rm mol[/tex].
Calculate the volume of a [tex]12.0\; \rm M[/tex] [tex]\rm NaOH[/tex] with that many [tex]\rm NaOH\![/tex] formula units:
[tex]\displaystyle V = \frac{n}{c} = \frac{12.5\;\rm mol}{12.0\; \rm mol \cdot L^{-1}}\approx 1.04\;\rm L[/tex].
That should be the volume of the [tex]12.0\; \rm M[/tex] [tex]\rm NaOH[/tex] solution needed to prepare that [tex]5.00\; \rm L[/tex] of [tex]2.50\; \rm M[/tex] [tex]\rm NaOH\![/tex] solution. However, [tex]1.04\; \rm L[/tex] corresponds to only about one-fifth the volume of a [tex]5.00\; \rm L\![/tex] solution. The difference in volume should be filled with pure water:
[tex]\begin{aligned}V(\text{water}) &\approx 5.00\; \rm L - 1.04\; \rm L = 3.96\; \rm L\end{aligned}[/tex].
