Plz solve this and explain it in detail.

Solution,
For the parallel combination of cells, effective emf is E and effective internal resistance is $\frac{r\times r}{r+r}=\frac{r}{2}$.
This combination is in series with another cell of emf E and internal resistance, r.
Thus, the net emf of this mixed combination of 3 cells, E_o = E + E = 2E.
and
the equivalent internal resistance of this mixed combination of cells, $r_o=\frac{r}{2}+r=\frac{3r}{2}$
Thus, current across the load R is
$i=\frac{E_o}{r_o+R}=\frac{2E}{{\displaystyle \raisebox{1ex}{$R+3r$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{4E}{2R+3r}$
Power dissipated by the resistor,
$$\begin{gathered} P=i^{2}R={\left(\frac{4E}{2R+3r}\right)}^{2}R \\ The power dissipated will be maximum when, \\ \frac{dP}{dR}=0 \\ Thus, \\ \frac{d}{dR}\left[{\left(\frac{4E}{2R+3r}\right)}^{2}R\right]=0 \\ 16E^2\frac{d}{dR}\left[{\left(\frac{1}{2R+3r}\right)}^{2}R\right]=0 \\ 16E^2 cannot be 0, thus \\ \frac{d}{dR}\left[{\left(\frac{1}{2R+3r}\right)}^{2}R\right]=0 \\ \frac{d}{dR}\left[\frac{R}{{\left(2R+3r\right)}^2}\right]=0 \\ \frac{{\left(2R+3r\right)}^2.1-R.2\left(2R+3r\right).2}{{\left(2R+3r\right)}^4}=0 \\ {\left(2R+3r\right)}^2.1-R.2\left(2R+3r\right).2=0 \\ {\left(2R+3r\right)}^2=4R\left(2R+3r\right) \\ \left(2R+3r\right)=4R \\ 2R=3r \\ R=\frac{3r}{2} \end{gathered}$$
Thus,
Ans (3)