It can be shown that there is only one strategy to handle the given problem hence no question of optimal strategy . Thats why they have asked for "number of stones" and not "minimum or maximum number of stones"..
Let we have (2n+1) number of stones so we have 'n' stones on one side of the middle stone and 'n' on other side
Distance between middle stone and right to it is 10 m..So to bring it to middle 20 m is covered..
Similarly for 2nd next one , we need distance = 40 m
and so on till nth next which is last from right , we need distance = 20n
Hence total distance = 20(1+2+3...+n) = 20n(n+1)/2 = 10n(n+1)
Likewise to bring the left stones we need total distance = 10n(n+1) m
Thus total distance = 20n(n+1) m which is equal to 4800 m as mentioned in the question.
20n(n+1) = 4800
==> n(n+1) = 240
==> n = 15
Hence number of stones = 2n + 1 = 2(15) + 1 = 31
Hence D) should be the correct option.