Given is an unbalanced Wheatstones' bridge, in which current distribution will be as shown



Now in loop $A B C A$, using KVL we have,
$$
\begin{aligned}
& -3 R i_1-4 R\left(i_1-i_2\right)+2 R i_2=0 \\
& -3 R i_1-4 R i_1+4 R i_2+2 R i_2=0 \\
& 7 i_1=6 i_2 \\
& \Rightarrow \\
& i_1=\frac{6}{7} i_2 \\
&
\end{aligned}
$$
Also, $\quad i_1+i_2=i$
Hence, $\left(\frac{6}{7}+1\right) i_2=i$ or $i_2=\frac{7}{13} i$
$$
\therefore \quad i_1=\frac{6}{7} i_2=\frac{6}{7} \times \frac{7}{13} i \Rightarrow i_1=\frac{6}{13} i
$$
Now from loop $A B D A$ (which includes cell), we have, $3 R i_1+2 R i_2 \approx V$ (nearly because internal resistance in not taken in account)
$$
3 R \times \frac{6}{13} i+2 R \times \frac{7}{13} i \approx i R_{\mathrm{eq}}
$$

$$
\Rightarrow \quad R_{\mathrm{eq}}=R\left(\frac{18+14}{13}\right)=\frac{32}{13} R
$$
For maximum power delivered to circuit, (internal resistance of source $)=($ external resistance)
$$
\Rightarrow \quad 5=\frac{32 R}{13} \text { or } R=\frac{13 \times 5}{32}=203 \Omega
$$
or $R \approx 2 \Omega$