To get maximum current in a resistance of 3ohm, one can use n rows of m cells (connected in series) connected in parallel.  If the total no. of cells is 24 and the internal resistance is 0.5 ohm then

m = 12 , n= 2 m=8 , n= 3 m = 2 , n = 12 m = 6 , n = 4

m = 12 , n= 2

It is given that m.n = 24 and for maximum current $\frac{mr}{n} = R$ $ \Rightarrow \frac{0.5 m}{n} = 3$ $ \frac{12}{n^2} = 3$ $ n^2 = 4, n = 2 , m = 12$