Comparing the pKb of piperazine and methylpiperazine
In terms of the logic employed by your answer key

Whether an inductive effect is stabilizing depends on the nature of the molecule and reaction you're thinking about.

You're right to think of the methyl group as inductively release electron density into the ring. However, this inductive release from the methyl group is actually increasing the electron density in the piperazine ring, which in turn makes the N lone pair higher in energy and more reactive, thus making it a stronger base. However!

What is actually the case in real life

I was able to find a detailed study of $\mathrm{p}K_\mathrm{a}$ values of protonated piperazines and methylpiperazines, and assuming $\mathrm{p}K_\mathrm{b} = 14 - \mathrm{p}K_\mathrm{a}$.

Then at 298 K...


$\mathrm{p}K_\mathrm{b}\text{(piperazine)} = 4.27$
$\mathrm{p}K_\mathrm{b}\text{(N-methylpiperazine)} = 4.86$
$\mathrm{p}K_\mathrm{b}\text{(N,N'-dimethylpiperazine)} = 5.62$
$\mathrm{p}K_\mathrm{b}\text{(2-methylpiperazine)} = 4.43$


This tells us that your answer key is actually wrong, and that in real life sterics may be playing a role in the basicity of the N atom. This data is in line with the fact that secondary bases tend to be slightly stronger than tertiary ones. For example, the 


$\mathrm{p}K_\mathrm{b}(\text{diethylamine)} = 3.02$
$\mathrm{p}K_\mathrm{b}(\text{triethylamine)} = 3.25$