Step 1: Power rule for determinants.
For any square matrix, $\det(A^{10})=(\det A)^{10}$, so the given condition becomes $(\det A)^{10}=1024$.
Step 2: Recognize $1024$.
Since $1024=2^{10}$, taking tenth roots gives $\det A=\pm 2$.
Step 3: Determinant of the $2\times 2$ matrix.
For $A=\begin{bmatrix}\alpha^2&5\\5&-\alpha\end{bmatrix}$, $\det A=\alpha^2(-\alpha)-5\cdot 5=-\alpha^3-25$.
Step 4: Solve $\det A=2$.
Then $-\alpha^3-25=2$ gives $\alpha^3=-27$, so $\alpha=-3$, which is one of the options.
Step 5: Check the other sign.
$\det A=-2$ gives $\alpha^3=-23$, which has no value matching the choices, so we discard it.
Step 6: Verify.
With $\alpha=-3$, $\det A=-(-27)-25=27-25=2$, and $2^{10}=1024$ as required.
\[ \boxed{\alpha=-3} \]