The correct answer is (d) z^3
For explanation: Given x(n) = δ(n+3)
We know that δ(n+3) = \(
\begin{cases}
1 &\text{\(n=-3\)} \\
0 &\text{otherwise} \\
\end{cases}\)
X(Z) = \(\sum\limits_{n=-\infty}^{\infty} x(n) z^{-n} = \sum\limits_{n=-\infty}^{\infty} δ(n+3) z^{-n}\) = z^3.