$$Let\ speed\ of\ correct\ hour\ =H$$
If mark resets four watches to 12 noon, what time will watch 4 get to show 12 am.
After every 1 hour watch 1 lose 15 min
Speed of watch 1 $$=\left(\frac{1}{1}-\frac{1}{4}\right)\cdot S=\frac{3}{4}\cdot S$$
Speed of watch 2 relative to watch 1 when watch 2 gains 15 min every hour ;
$$\frac{75}{60}\cdot\frac{3}{4}\cdot S$$
$$=\frac{5}{4}\cdot\frac{3}{4}\cdot S$$
Speed of watch 3 relative to watch 2 when watch 3 loses 20 min every hour ;
$$\left(\frac{1}{1}-\frac{20}{60}\right)\cdot\frac{5}{4}\cdot\frac{3}{4}\cdot S$$
$$\frac{4}{3}\cdot\frac{5}{8}\cdot S=\frac{5}{8}\cdot S$$
Speed of watch 4 relative to watch 3 when watch 4 gains 20 minutes every hour ;
$$\frac{80}{60}\cdot\frac{5}{8}\cdot S$$
$$=\frac{4}{3}\cdot\frac{5}{8}\cdot S=\frac{20}{24}\cdot S=\frac{5}{6}\cdot S$$
i.e watch 4 will lose $$\left(\frac{1}{1}-\frac{5}{6}\right)=\frac{1}{6}of\ \ 1hour\ for\ every\ hour$$
Total loss in 12 hours $$=\frac{1}{6}\cdot\frac{12}{1}=\frac{12}{6}=\ 2\ hours$$
By 12:00 am (midnight) watch could have lost 2 hours, hence it will show (12-2 = 10 hours)
$$Answer\ is\ option\ A$$