Let the independent random variables $X_1$, $X_2$, $X_3$ have the same cdf $F(X).$ Let $Y$ be the middle value (second largest) of $X_1$, $X_2$, $X_3$. Determine the cdf of $Y$. (Hint: use a Binomial distribution).
My approach is that this is a binomial distribution where I am trying to calculate the probability that $2$ of that random variables are less than or equal to some number $x$ and one is of the random variables is larger. Is this the right approach?
That is correct. $$ F_{\text{middle}} (x) = \Pr(\text{the middle one}\le x) = \Pr(\text{number of successes}\ge 2) $$ where "success" on the $i$th trial means $X_i\le x$, and that has a probability that depends on $x$.