For $R_3=0,I=9 \mathrm{A}$ (from the graph).In this situation, the circuit can be drawn as shown in figure $\left(\mathrm{A}\right)$ .



[current through R₂ will be zero because we have a short circuit $\left(R_3=0\right)$ across $R_2$]

Here $I=\frac{36}{R_1+R_2}=9 \mathrm{A}$

or $R_1+R_2=4 \mathrm{\Omega }$ ...(1)

current will not flow through $R_3$, it will be open circuit.

In this situation, the circuit can be drawn as shown figure $\left(\mathrm{B}\right)$ .


Figure (B)

Now $I=\frac{36}{R_1+2R_{\mathrm{2 }}}=6 \mathrm{A}$ (given)

$\therefore \mathrm{ }{R}_1+2R_2=6 \mathrm{\Omega }$ ...(2)

from Eqs. (1) and (2)

$R_1=2 \mathrm{\Omega },\mathrm{ }{R}_2=2 \mathrm{\Omega }$