I will assume that "we are shown one bead" means that we are shown one bead, chosen at random from the beads in the selected envelope.
Since intuition can sometimes lead us astray, we will use the formal conditional probability machinery. Let $W$ (for "win") be the event that the envelope contains $2$ red and $2$ black, and let $B$ be the event that the ball that was drawn is black. We want the conditional probability $P(W|B)$.
By the definition of conditional probability, we have $P(W|B)=\frac {P(W\cap B)}{P(B)}.$
We compute the two required probabilities on the right.
The probability of $W\cap B$ is $(1/2)(2/4)$. This is because we must pick the correct envelope (probability $1/2$) and draw one of the $2$ black beads from the $4$ beads in the envelope.
The probability of $B$ is $(1/2)(2/4)+(1/2)(2/3)$. This is because the event "we get a black bead" can happen in two disjoint ways: (i) We pick the envelope with $2$ black, $2$ red and draw a black bead or (ii) We pick the envelope with $2$ black, $1$ red and draw a black bead.
Divide and simplify. We get $3/7$.
The expectation of the amount of money in the envelope is therefore $(3/7)(1)+(4/7)(0)$, which is $3/7$.
As to how much one "should" pay for the envelope, the commonsense answer is as little as possible. But if we pay $3/7$ of a dollar for the envelope, then our expected total gain is $0$, making it a "fair" game. That is undoubtedly the expected answer.
A Somewhat Different Analysis: Let $a$ be the amount that we pay for the envelope, and let $X$ be our net gain.
If there is a dollar in the envelope, the net gain is $1-a$. If there is no money in the envelope, our net gain is $-a$. Thus $E(X)=(1-a)(3/7)+(-a)(4/7).$ Simplify. We get $E(X)=\frac{3-7a}{7}.$ This expectation is $0$ ("fair game") if $3-7a=0$, that is, if $a=3/7$.
Reality Check: The envelope with no money is (proportionally) richer in black than the envelope with money. So drawing a black bead means that there is a better than $1/2$ probability that we are dealing with the no money envelope. Thus the expected value of the money in the envelope is less than $50$ cents. And indeed $3/7$ is less than $1/2$. It is almost always a good idea to check whether a computed answer "makes sense."
Comment: If the bead we are shown is not chosen at random, the answer depends on the strategy that the opponent is using. For example, if the opponent always shows you a black bead, whatever envelope you choose, then the probability of winning a dollar is $1/2$, so a payment of $50$ cents makes the game fair.