Step 1: Locate where the water line meets the upstream face.
Dam height = $12+2=14$ m (water level plus freeboard). On the $2:1$ upstream slope, the $12$ m water line sits $2\times12=24$ m from the heel.
Step 2: Find the base width and the filter's near edge.
Upstream face run $=2\times14=28$ m, downstream face run $=2.5\times14=35$ m, so the base spans $28+3+35=66$ m. The $20$ m filter, laid in from the toe, starts $66-20=46$ m from the heel.
Step 3: Correct the entry point for the curved seepage line.
The real seepage surface does not leave the reservoir straight from $A$; the equivalent parabola starts closer to the dam, at $0.3\times24=7.2$ m in from $A$, i.e. at $24-7.2=16.8$ m from the heel, with the full head $h=12$ m still acting there.
Step 4: Set up the focus-to-entry distance and solve.
$L = 46-16.8 = 29.2$ m. Treating the flat filter as the focus of a Dupuit-type base parabola, the flow it collects per metre length is
\[ q = k\left(\sqrt{L^2+h^2}-L\right) \]
\[ q = 2.5\times10^{-5}\left(\sqrt{29.2^2+12^2}-29.2\right) \]
\[ \sqrt{852.64+144} = \sqrt{996.64} = 31.57 \]
\[ q = 2.5\times10^{-5}(31.57-29.2) = 2.5\times10^{-5}\times2.37 \]
Final Answer:
This gives $q \approx 5.92\times10^{-5}$ m$^3$/s per metre length of the embankment.
\[ \boxed{n \approx 5.92} \]