Right answer is (d) \frac{a}{z(z-a)^2}
To explain: Given x(n) = n[a^n u(n)]
We know that a^n u(n) ↔ \frac{z}{z-a}
Time differentiation property states that
If x(n) ↔ X(z), then nx(n) ↔ -z \frac{dX(z)}{dz}
Z[x(n)] = Z{n[a^n u(n)]} = -z \frac{dX(z)}{dz} = -z \frac{d}{dz} \Big[\frac{z}{z-a}\Big] = \frac{a}{z(z-a)^2}.