if his investment were to earn an interest £800, then his accumulated amount will be 300 + 800 or namely £3800.
let's check at what point he'll be getting an accumulated amount of £3800, and then we know that compounding it further will simply result in more bucks, thus we know that'd be the least it'd take, keeping it further will simply give more.


so depends, I guess if he just waits one more minute that may do it =).