Let $\langle M \rangle$ be the encoding of a Turing machine as a string over $\Sigma = \{0,1\}$. Let $L = \{\langle M \rangle \mid M\text{ is a Turing machine that accepts a string of length 2014}\}$. Then, $L$ is

1. Undecidability (Rice's Theorem): The property "$M$ accepts at least one string of length 2014" is a non-trivial property of the language recognized by $M$, $L(M)$:

   • There exists a Turing machine $M_1$ that accepts $\{0,1\}^{2014}$, so $\langle M_1 \rangle \in L$.

   • There exists a Turing machine $M_2$ with $L(M_2) = \emptyset$, so $\langle M_2 \rangle \notin L$.

   By Rice's Theorem, testing non-trivial semantic properties of Turing machine languages is undecidable.

2. Turing Recognizability / Recursive Enumerability: We can construct a Turing machine $M^{\prime}$ that recognizes $L$ using dovetailed simulation:

   • The set of strings of length 2014 is finite: there are $2^{2014}$ such strings, say $w_1, w_2, \dots, w_k$.

   • For $s = 1, 2, 3, \dots$, simulate $M$ on each string $w_i$ for $s$ steps.

   • If $M$ accepts any string $w_i$, $M^{\prime}$ halts and accepts $\langle M \rangle$.

   • If $M$ never accepts any string of length 2014, $M^{\prime}$ runs forever.

   Thus, $L$ is semi-decidable (recursively enumerable).

Hence, $L$ is undecidable but recursively enumerable, corresponding to option (B).