jmc

From $1$ to $101$ there are $\lfloor \frac{101}{2}\rfloor =50$ (see floor function) multiples of $2$, and their average is $$\begin{gathered} \frac{2\cdot 1+2\cdot 2+2\cdot 3+\cdots +2\cdot 50}{50} \\ =\frac{2(1+2+3+\cdots +50)}{50} \\ =\frac{2\cdot \frac{50\cdot 51}{2}}{50} \\ =\frac{2\cdot 51}{2} \\ =51 \end{gathered}$$ Similarly, we can find that the average of the multiples of $3$ between $1$ and $101$ is $51$, the average of the multiples of $4$ is $52$, the average of the multiples of $5$ is $52.5$, and the average of the multiples of $6$ is $51$, so the one with the largest average is $\boxed{\text{multiples of 5 between 1 and 101}}$