Solution idea
D=Demand= 24000
S=Ordering/Setup cost= 2400
H=Holding/Carrying Cost= 3
U= Usage rate=24000/240= 100
P= Production rate= 600

Optimal Run Size:
Q0= √2DS/H√p/(p-u)
Q0= √(2240002400)/3√600/(600-100)
Q0= √115200000/3√600/500
Q0= √38400000* √1.20
Q0= 6196.77*1.0954
Q0= 6787.94 ≅6788

Minimum total annual cost for carrying and setup cost.
= Carrying Cost + Set up Cost
=( I max/2)H+ ( D/Q0)S
Where,
I max= Q0/p ((p-u))
I max= 6788/600 (600-100)
I max= 11.31 (500)
I max= 5655
TC min= Carrying Cost+ Setup Cost
TC min= (I max/2) H+ (D/Q0) S
TC min= (5655/2) 3 + (24000/6788) 2400
TC min= (2827.5) (3) + (3.53) (2400)
TC min= 8482.5 + 8472
TC min= 16954.5

Cycle time for the Optimal Run Size.
Q0/U=6788/100
= 67.88 ≅ 68 days

Run time
Q0/p=6788/600
= 11.31 days