Since you are just looking for an approximation, we can just compute the arc length subtended by the angle.
The arc subtended by angle $\theta$ (in radians) at a radius $r$ has length $r\theta$.
Hence the diameter is $\approx 5.6 \times 10^7 \cdot 0.00012 \approx 6720$ km.
A more exact answer, which notes that the extreme rays touch Mars tangentially, shows that the diameter is $2 r \tan {\theta \over 2} \approx 6720$ km.