In the given Wheatstone Bridge no current flows through $15 \Omega$ resistance. The value of the unknown resistance 'R' is:

$16 \Omega$

The correct answer is Option (4) → $16 \Omega$

In the given wheatstone bridge no current flows through 15Ω.

$⇒\frac{4}{R_4}=\frac{2}{4}⇒R_4=\frac{4×4}{2}=8Ω$

and,

$\frac{1}{R_4}=\frac{1}{16}+\frac{1}{R}$ [connected in parallel]

$\frac{1}{R_4}=\frac{1}{8}-\frac{1}{16}=\frac{2-1}{16}$

$R=16 \Omega$