As the others answered, it's not the ISS, but I wanted to answer about the posibility of it being the ISS anyway.
At first I thought it can't be the ISS, because I like to observe the ISS at night and it's moving much faster than in your picture, but then I made some more precise calculations.
Your 18mm lens on a crop body has a 66°×47°; 76° (H×V; D) field of view. Your trail is 68 pixels long, and your image diagonal is 961 pixels, so your trail occupies
68/961 × 76° = 5.37°.
At 30s exposure this means an average angular velocity of 0.179°/s.
Your trail is approximately 215 pixels over the horizon, and the image is 533 pixels tall, so
215/533 × 47° = 18.9° over the horizon. Let's assume the ISS is
α=18.9° over the horizon relative to your position. Assuming the height of the ISS
(h = 405km) is much smaller than the earth radius, some basic trigonometry put its distance relative to you to
D = h/sin α = 1253km. Now its observed angular velocity depends on its motion relative to you, but let's go for the average scalar velocity, which is
ωₐ = 1/(2π) * Integral[ω*|sin[x]|, 0, 2π], where
ω is the best case (tangential motion) velocity.
ω = v/D, where
v = 7.76km/s, the orbital speed of the ISS, so
ω = 6.19 × 10⁻³ rad/s. Evaluating the integral we get
ωₐ = 3.94 × 10⁻³ rad/s. In 30s it would have moved `0.118 rad = 6.76°.
This number is comparable to your trail, which surprised me, because I know the ISS can move much faster. But of course, I observe it in the best circumstances, when it's directly above. If the ISS is directly above,
ω = 19.16 × 10⁻³ rad/s = 1.097°/s.