3. Choose D.

Brief solution: From the given $2^{5}=\sqrt[11]{a_{1} \cdot a_{1} q \cdot a_{1} q^{2} \cdots \cdot a_{1} q^{10}}=\sqrt[11]{a_{1}^{11} q^{\frac{10(1)+1)}{2}}}=\sqrt[11]{a_{1}^{11} q^{55}}=a_{1} q^{5}$.
Since $a_{1}=2^{-5}$. Therefore, $q=2^{2}=4$. Let the term removed be $a_{n}=a_{1} q^{n-1}$, then $2^{4}=\sqrt[10]{\frac{a_{1}^{11} q^{55}}{a_{1} q^{n-1}}}=$ $\sqrt[10]{a_{1}^{10} q^{36-n}}$.
Thus, $n=56-45=11$, which means the 11th term is removed.
3. For a geometric sequence $\left\{a_{n}\right\}$, the first term $a_{1}=2^{-5}$, the geometric mean of its first 11 terms is $2^{5}$, and if one term is removed from the first 11 terms, the geometric mean of the remaining terms is $2^{4}$. The term removed is
A. $a_{8}$
B. $a_{9}$
C. $a_{10}$
D. $a_{11}$