How many possible panels have two women and two men? That would be $\displaystyle {_{17}C_2}\cdot {_{18}C_2}$. So, your denominator is wrong.
Anyway, it is just asking about the women. The men don't factor into it at all. Given that you choose two women at random, what is the probability that they are both doctors. That would be $\displaystyle \dfrac{12\cdot 11}{17\cdot 16}\approx 0.49$ I dunno where 0.807 comes from.
The problem said assuming there are exactly two women on the panel, what is the probability that both are women. This is exactly the same as asking "If two women are selected from a group of 17, 12 of whom are doctors, what is the probability they are both doctors. The number of men is not relevant.
Had the question been worded, "If at least two women are chosen for the panel, what is the probability that at least two women doctors are chosen", then the answer would be:
$\displaystyle \dfrac{ {_{12}C_2}\cdot {_{23}C_2} + {_{12}C_3}\cdot {_{23}C_1} + {_{12}C_4}\cdot {_{23}C_0} }{ {_{17}C_2}\cdot {_{18}C_2} + {_{17}C_3}\cdot {_{18}C_1} + {_{17}C_4}\cdot {_{18}C_0} } = \dfrac{1309}{2084} \approx 0.628$
I don't think I know of a question similar to the one asked that could yield a probability as high as they want.