This is a good example of Bayes Rule: $$P(B_1|V)=\frac{P(V|B_1)P(B_1)}{P(V)}$$
Your calculation gives $P(V|B_1)P(B_1)=P(B_1\cap V)$, but the denominator needs to be calculated as well:
$$P(V)=P(V|B_1)P(B_1)+P(V|B_2)P(B_2)$$
A toy example: Bowl 1 contains 1 vanilla, Bowl 2 contains 1 chocolate. Say, we select a random bowl and select a random cookie. If the chosen cookie, is vanilla, what is the probability that it's chosen from Bowl 1?
Your way gives $1/2\times1=1/2$, but actually it's certain, i.e. probability is 1, that the cookie is from Bowl 1. Keep in mind that what you're asked is $P(B_1|V)$, not $P(B_1\cap V)$.
Note: $V|B_1$ is not an event. This notation makes sense inside $P(.)$ expression. So, you should define your events carefully and express the question in terms of these events.