Yes, it is just $\operatorname{Sym}^k(V)$ itself.
Specifically: Let $V = \mathbb{C}^{2n}$ be the natural module for $Sp(2n, \mathbb{C})$. Then for all $1 \leq k \leq n$, the symmetric power $\operatorname{Sym}^k(V)$ is an irreducible $Sp(2n,\mathbb{C})$-module of highest weight $(k,0,\ldots,0) = k \varpi_1$.
See for example §24.2, p. 406 in Fulton-Harris.