Three resistances each of $4\mathrm{\Omega }$ are connected to form a triangle. The resistance between any two terminals is 

The two resistances are connected in series and the resultant is connected in parallel with the third resistance. 

$\therefore R^{\mathrm{\prime }}=4\mathrm{\Omega }+4\mathrm{\Omega }=8\mathrm{\Omega }\text{ and }\frac{1}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$

 $\Rightarrow {\mathrm{R}}_{\mathrm{eq}}=\frac{8}{3}\mathrm{\Omega }$