The two geometric means between the numbers 1 and 64 are

(b) 4 and 16

Let the two G.M.s between 1 and 64 be $G_{1}$ and $G_{2}$.

Thus, $1, G_{1}, G_{2}$ and 64 are in G.P.

$64=1 \times r^{3}$

$\Rightarrow r=\sqrt[3]{64}$

$\Rightarrow r=4$

$\Rightarrow G_{1}=a r=1 \times 4=4$

And, $G_{2}=a r^{2}=1 \times 4^{2}=16$

Thus, 4 and 16 are the required G.M.s.