The information that you are given is about the behavior of the number of spam emails, GIVEN the total number of emails. Although it isn't explicitly spelled out, if $N$ is the (stochastic) number of emails received, then the number of spam emails follows the Binomial distribution with $N$ trials and probability of "success" (spam) $p$.
How to use this? The law of total probability says that if $K$ is the number of spam messages, then
$$
P(K=k)=\sum_{n=0}^{\infty}P(K=k\mid N=n)\cdot P(N=n).
$$
We already know that
$$
P(N=n)=e^{-200}\frac{200^n}{n!}.
$$
And, condition on $N=n$, we know the distribution of $K$: in particular, it is $\text{Bin}(n,p)$. You can use this to compute, for any $n$, $P(K=k\mid N=n)$. From there, it is just a matter of completing the summation.
Let $X$ represent the number of emails you receive in an hour on a weekday and let $Y$ represent the number of emails you receive in an hour on a weekend. Let $Z$ be the random variable representing how many emails you received in an hour knowing that the day you selected could be any day of the week.
$$X\sim\mathrm{Po}(10)\implies\operatorname{P}(X=x)=\frac{10^x}{e^{10}x!}\tag{1}$$
$$Y\sim\mathrm{Po}(2)\implies\operatorname{P}(Y=y)=\frac{2^y}{e^{2}y!}\tag{2}$$
Bayes’ theorem tells us
$$\operatorname{P}(B \,|\, A) = \frac{ \operatorname{P}(B) \, \operatorname{P}(A \,|\, B) }{\operatorname{P}(B) \, \operatorname{P}(A \,|\, B) + \operatorname{P}\left(B'\right) \, \operatorname{P}\left(A \,\middle|\, B'\right)}$$
Taking $B$ to be the event $D$ and $A$ to be the event $Z=0$, we have
$$\operatorname{P}(D \,|\, Z=0) = \frac{ \operatorname{P}(D) \, \operatorname{P}(Z=0 \,|\, D) }{\operatorname{P}(D) \, \operatorname{P}(Z=0 \,|\, D) + \operatorname{P}\left(D'\right) \, \operatorname{P}\left(Z=0 \,\middle|\, D'\right)}$$
Given if $D$ occurs, $Z=0$ can actually be more precisely written as $X=0$; if $D'$ occurs, then $Z=0$ can actually be more precisely written as $Y=0$, giving you
$$\operatorname{P}(D \,|\, Z=0) = \frac{ \operatorname{P}(D) \, \operatorname{P}(X=0 \,|\, D) }{\operatorname{P}(D) \, \operatorname{P}(X=0 \,|\, D) + \operatorname{P}\left(D'\right) \, \operatorname{P}\left(Y=0 \,\middle|\, D'\right)}$$
By the definition of our variables, we can further contract this to
$$\operatorname{P}(D \,|\, Z=0) = \frac{ \operatorname{P}(D) \, \operatorname{P}(X=0) }{\operatorname{P}(D) \, \operatorname{P}(X=0) + \operatorname{P}\left(D'\right) \, \operatorname{P}(Y=0)}$$
because if we’re analyzing $X$, we know we’re looking at a weekday, and if we’re analyzing $Y$, we know we’re looking at a weekend.
Use $(1)$ and $(2)$ in conjunction with $\operatorname{P}(D)=5/7$ and $\operatorname{P}\left(D'\right)=2/7$ to arrive at your answer.
Best Answer
The binomial distribution as it was used in the question is not correct unless the distribution of the number of emails received in a day is not random; i.e., exactly 25 emails are received each day, and the probability that any given email is spam is exactly 60%.
If, however, we assume that the number of emails received in a day is a random variable $N$ which is Poisson distributed with mean $\lambda = 25$, the conditional distribution of the number of spam emails received on a given day with $N = n$ total emails is $$X \mid N = n \sim \operatorname{Binomial}(n,0.6).$$ The unconditional distribution is therefore $$\Pr[X = x] = \sum_{n=x}^\infty \Pr[X = x \mid N = n]\Pr[N = n] = \sum_{n=x}^\infty \binom{n}{x} p^x (1-p)^{n-x} e^{-\lambda} \frac{\lambda^n}{n!} = e^{-p\lambda} \frac{(p \lambda)^x}{x!},$$ which is of course itself Poisson with rate parameter $p\lambda$. So the probability that $X = 15$ is simply $$\Pr[X = 15] \approx 0.102436.$$ In comparison, if $N = 25$ and we use the binomial model, $$\Pr[X = 15] \approx 0.161158.$$ (Your calculation is not correct because you've switched $p$ and $1-p$.)
Under a different distributional assumption for the number of emails received in a day, the answer will be different. $N$ need not be $25$, nor does it need to be Poisson. It could be negative binomially distributed; it could be uniform; it could be any discrete distribution whose support is a subset of the nonnegative integers. The question as it is posed does not impose such a distribution nor does it imply one--it only supposes that its mean is $25$.