radius = 10
$$Circumference=2\pi r=\pi\cdot2\cdot r=\pi\cdot2r$$
Where 2r = Diameter ;
$$C=\pi\left(20\right)or\ 20\pi$$
Car B leaves 10 hrs before car A hence Car B have travelled the distance 2*10 = 20 miles which is in 10 hrs before car A and B reach one another they will cover distance of
$$\left(\pi20-20\right)miles$$
Rate A + Rate B = 2 miles/hr + 3 miles/hr = 5 miles/hr
$$Time=\frac{dis\tan ce}{rate\ }=\frac{\left(\pi20-20\right)}{5}=\left(\pi4-4\right)hours$$
It takes the time above for them to reach one another, so to travel past one another with distance of 12 miles. It will be
Distance = rate * time
12 = 5t ; t=12/5 = 2.4hrs
Therefore,
Car B has been on the circuit for
$$\left(\pi4-4\right)+\left(2.4\right)+10$$
$$\left(\pi4-4+12.4\right)hrs$$
$$\left(\pi+8.4\right)hrs$$
$$answer\ is\ Option\ B$$