Method 1: The numerator should be $1/3$. HINT: What should be the probability that the maximum of the three numbers $\{ X_1, X_2, X_3 \}$ turns out to be $X_1$?
Method 2: Getting the same answer using integrals: Let $F(\cdot)$ be the c.d.f. of the distribution. Conditioned on $X_1 = x$, the probability that $X_2$ (resp. $X_3$) lies in the interval $(-\infty, x)$ is $F(x)$. Therefore, the desired probability is $ p = \int_{-\infty}^{+\infty} F(x)^2 F^{\ \prime}(x) dx. $ Using integration by parts, $ p = \left. F(x)^2 F(x) \right|_{-\infty}^{+\infty} - \int_{-\infty}^{\infty} 2 F(x) F^{\ \prime}(x) \times F(x) dx = \left. F(x)^3 \right|_{-\infty}^{+\infty} - 2p. $ Can you now simplify to get $p = 1/3$?
Method 3: Using triple integrals: Let $f(\cdot)$ denote the p.d.f. of the distribution. Then, the "happy" event whose probability we are computing is that $(X_1, X_2, X_3) \in D$ where $ D := \{ (x, y, z) \mid x \geq y, x \geq z \} \subseteq \mathbb R^3. $ So the probability is given by $ \iiint_D f_{X_1, X_2, X_3}(x,y,z) \ dx \ dy \ dz = \iiint_D f(x) f(y) f(z) \ dx \ dy \ dz. $ Now, fixing an $x \in \mathbb R$, we note that $y$ and $z$ lie in the interval $(-\infty, x)$. So the probability is given by $ \int_{-\infty}^\infty f(x) \left( \int_{-\infty}^x f(y) dy \right) \left( \int_{-\infty}^x f(z) dz \right) dx, $ which can be written equivalently as $ \int_{-\infty}^\infty f(x) F(x)^2 dx, $ confirming the integral in Method 2.
Another approach. The original problem can also be solved directly as follows. Sort the three numbers $X_1, X_2, X_3$ and rank them in decreasing order, so that $r_i$ is the rank of $X_i$. (We assume that $X_1, X_2, X_3$ are pairwise distinct; this assumption is ok if this fails to occur with zero probability.) Then since $(X_1, X_2, X_3)$ are iid, the random variable $r := (r_1, r_2, r_3)$ has the uniform distribution over the $3! = 6$ permutations of $(1, 2, 3)$.
Conditioned on $X_1 > X_3$, $r$ takes the uniform distribution over $\{(1,2,3), (1,3,2), (2,1,3)\}$. Of these three possibilities, two of them (namely, ${ (1, 2, 3), (1, 3, 2)}$) correspond to the event $X_1 > X_2$. Hence the probability that $X_1 > X_2$ conditioned on $X_1 > X_3$ is $\frac {2}{3}$.