Who's up to the challenge?17

By the definition of the Mangoldt Function we know that $\lambda(n)=log(p)$ if $n=p^k$ , where $p$ is a prime and $k \ge 1$ is a natural number and $0$ otherwise.

So we have $\sum_{d|n}^{} \lambda(d)=log(n)$ which follows from the Fundamental Theorem of Arithmetic .

Let $n={p_1}^{\alpha_1}*...*{p_t}^{\alpha_t}$ be the canonical decomposition of $n$ so that the $p_i$ 's are the primes and $\alpha_i$ 's are the powers of the primes.

Now $sum_{d|n}^{} \lambda(d)$ will be same as the sum of $\lambda({p_i}^{\beta_i})$ for all $0 \le \beta_i \le \alpha_i$ , so it will be equal to

${\alpha_1}*{log({p_1})}+{\alpha_2}*{log({p_2})}+...+{\alpha_t}*{log({p_t})}$

$=log({p_1}^{\alpha_1}*...*{p_t}^{\alpha_t})$

$=log(n)$

So the sum in the power of $e$ in the above expression is same as $log(n)$ .

Hence the answer is $n=896$ .