A term of the binomial expansion takes the form \[a_k = \binom{31}{k} \left(\frac{1}{2}\right)^k,\]where $ 0 \le k \le 31 $. To find which $ a_k $ is the largest, we evaluate the ratio $ \frac{a_{k+1}}{a_k} $: \[\frac{a_{k+1}}{a_k} = \frac{\binom{31}{k+1} \left(\frac12\right)^{k+1}}{\binom{31}{k} \left(\frac12\right)^k} = \frac{\frac{31!}{(k+1)!(30-k)!} \left(\frac12\right)^{k+1}}{\frac{31!}{k!(31-k!)} \left(\frac12\right)^k} = \frac{31-k}{2(k+1)}.\]Now, the inequality $ \frac{31-k}{2(k+1)} > 1 $ is equivalent to $ 31-k > 2k+2, $ or $ k < \frac{29}{3}, $ or $ k \le 9 $. Furthermore, $ \frac{31-k}{2(k+1)} < 1 $ when $ k > \frac{29}{3}, $ or $ k \ge 10 $. Therefore, $ a_{k+1} > a_k $ for $ k \le 9 $ and $ a_{k+1} < a_k $ for $ k \ge 10 $. It follows that $ a_{10} $ is the largest term of the binomial expansion. We have \[a_{10} = \binom{31}{10} \left(\frac12\right)^{10},\]so it suffices to find the power of $ 2 $ in the prime factorization of $ \binom{31}{10} $. We have \[\binom{31}{10} = \frac{31 \cdot 30 \cdot 29 \cdot 28 \cdot 27 \cdot 26 \cdot 25 \cdot 24 \cdot 23 \cdot 22 \cdot 21}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{A \cdot 2^{8}}{B \cdot 2^8} = \frac{A}{B},\]where $ A $ and $ B $ are odd integers. Therefore, $ \binom{31}{10} $ is odd, and so the answer is $ 2^{10} = \boxed{1024} $.