Ans is an option (C)
Let the diameter of the track be $d$ mtr. Then the length of the track is $\pi d$ mtr. Let the speeds of ant$_{1}$ and ant$_{2}$ be $v_{1},v_{2}$ mtr/min respectively.
Then, let in time $t$ minutes, ant$_{1}$ covers $60$% of the track i.e. distance of $0.6\pi d$ mtr.
Then in the same time $t$ minutes, ant$_{2}$ covers $40$% of the track i.e. distance of $0.4\pi d$ mtr.
For ant$_{1}$ we have following equations: $t=\frac{0.6\pi d}{v_{1}} \longrightarrow(1)$ and $t+12=\frac{\pi d}{v_{1}}\longrightarrow(1)$
Dividing $(2)$ by $(1),$ we get $t=18$ minutes
For ant$_{2}$ we have following equations: $t=\frac{0.4\pi d}{v_{2}}\longrightarrow(3)$ and $t+y=\frac{\pi d}{v_{2}}\longrightarrow(4) \; (t+y$ is the complete time taken by ant$_{2}$ to complete the track) .
Dividing $(4)$ by $(3),$ we get $y=27$ minutes
After $t$ minutes, it is $10:00$ A.M when the two ants meet for the first time. From there, it takes $27$ minutes for ant$_{2}$ to reach point P.
So, ant$_{2}$ reaches point P at $10:27\;\text{A.M.}$