Correct option is (d) \(\frac{4}{z(z-2)}\)
For explanation I would say: Given x(n) = 2^n u(n-2)
Time shifting property of Z-transform states that
If x(n) ↔ X(z), then x(n-m) ↔ z^-m X(z)
Z[u(n-2)] = z^-2 Z[u(n)] = \(z^{-2} \frac{z}{z-1} = \frac{1}{z(z-1)}\)
The multiplication by an exponential sequence property of Z-transform states that
If x(n) ↔ X(z), then a^n x(n) ↔ X(z/a)
Z[2^n u(n-2)] = Z[u(n-2)]|z=(z/2) = \(\Big[\frac{1}{z(z-1)}\Big]_{z=(z/2)}\)
\( = \frac{1}{(z/2)[(z/2)-1]} = \frac{4}{z(z-2)}\).