Answer: B

Small Hint:

If the last written number is $n,$ the remaining count is $n-1$

Big Hint:

Use the bounds $1\le e\le n$ on the erased value to narrow the possible values of $n$

Solution:

The new average $\frac{602}{17}$ is below the original average $\frac{n+1}{2}$ but differs from it by less than $\frac{n}{n-1},$ forcing $n=69$ or $70.$ Since the remaining sum $(n-1)(\frac{602}{17})$ must be integral, $n-1$ is divisible by $17,$ so $n=69.$ The erased number is $$\begin{aligned} 2415-68\cdot\frac{602}{17} &=2415-2408\\ &=7. \end{aligned}$$

Therefore, the correct answer is B.