Here are a couple of ways to go about making some informed decisions in estimating a line integral given a picture:
We note the path is a simple path, it is a semi-circle of radius $r=1$ and total arc length of $\pi$, this can help in the estimation step. We also note that $f$ is approximately monotone decreasing along the path in such a way as to make estimation of values between the contour lines somewhat simple or at least probable.
First lets use a trick from single variable calculus to constrain our integral between its max and min values. We will take two integrals, both treating $f$ as a constant set by the beginning and ending of the path (because $f$ is approximately monotone we have a highest and lowest point). I will take these points as $f(3,3)\approx 4.1$ the beginning of the curve above the $f=4$ contour, and $f(3,1)\approx 1.3$ the point on the curve above the $f = 1$ contour. We then compute the two approximate integrals to help us bound the answers a bit.
$$\int_C f(x,y)ds = \int_C f(3,3)ds \approx \int_C (4.1)ds = 4.1 \cdot \pi \approx 12.88 $$
$$\int_C f(x,y)ds = \int_C f(3,1)ds \approx \int_C (1.3)ds = 1.3 \cdot \pi \approx 4.08 $$
At least we have cut off anything less in magnitude than $4$, so the answers $-3,0,3$ are all too small.
To decide if the answer is closer to $9$ or $6$, I appeal to the drawing, the path spends more time traveling through the segment (is longer)
$3<f<4$ than the segment $2<f<3$, this pushes the value closer to $9$. And lastly the integration path travels backwards towards decreasing values of $f$ and this could make the value of the integral negative, I would suspect that the value of the integral is most likely $\pm9$ with a preference towards $-9$