There are $48$ cells, each of $emf 2V$ and internal resistance $1 \Omega .$ For maximum current through an external resistance of $3 \Omega ,$ they should be connected as:

Let us connect m cells in series and n such groups in parallel, where $mn = 48.$ 

Current in the mixed grouping 

$$\begin{gathered} I=\frac{mE}{\left({\displaystyle \raisebox{1ex}{$mr$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}\right)+R} \\ for maximum current \frac{mr}{n}=R \\ \frac{m}{n}=\frac{R}{r}=\frac{3}{1} \\ Solving, we get \\ m=12 and n=4 \end{gathered}$$