Adding or removing one of repeating member does not alters the resistance of an infinite network. Let $R_{A B}=x$, then $R_{A B}=R_{C D^*}$

$\Rightarrow \quad x=r+\frac{r x}{r+x}$

$\Rightarrow \quad x^2-r x-x=0$

From sridharacharya formula, we have

$x=-b \pm \frac{\sqrt{b^2-4 a c}}{2 a}$

$\Rightarrow x=-\frac{(-r) \pm \frac{\sqrt{r^2}}{2}+4 r^2}{2}$

$\Rightarrow x=\frac{r(1+\sqrt{5})}{2}$

Now, power consumed by bulb of resistance $R$ is $1 \,W$,

$i^2 R=1 \Rightarrow i^2=\frac{1}{16} \Rightarrow i=\frac{1}{4} \,A$

Now, current in circuit is

$i=\frac{V}{R_{\text {total }}} \Rightarrow i=\frac{V}{R+R_{C D}}$

$\Rightarrow \frac{1}{4}=\frac{10}{16+\frac{r}{2}(1+\sqrt{5})}$

$\Rightarrow 16+\frac{r}{2}(1+\sqrt{5})=40 \Rightarrow r=14.8 \,\Omega$