Right through the center

The hole is basically a cylinder with two spherical caps on each end. The following picture shows a cross-section of the sphere with the hole:

As given, the radius of the sphere is $R = OA = OC = OD$ , and the radius of the cylinder is $\frac{1}{2}R = AB = BC$ . By the Pythagorean Theorem on $\triangle ABO$ , $OB = \frac{\sqrt{3}}{2}R$ , which means $BD = OD - OB = R - \frac{\sqrt{3}}{2}R = \frac{2 - \sqrt{3}}{2}R$ .

The volume of the cylinder is then $V_{\text{cyl}} = \pi r^2h = \pi \cdot (\frac{1}{2}R)^2 \cdot 2 \cdot \frac{\sqrt{3}}{2}R = \frac{\sqrt{3}}{4}\pi R^3$ .

And the volume of one spherical cap is $V_{\text{cap}} = \frac{1}{3}\pi h^2(3R - h) = \frac{1}{3}\pi (\frac{2 - \sqrt{3}}{2}R)^2(3R - (\frac{2 - \sqrt{3}}{2}R)) = \frac{16 - 9\sqrt{3}}{24}\pi R^3$ .

The percent of the sphere volume that has been drilled out is therefore $p = \frac{V_{\text{cyl}} + 2V_{\text{cap}}}{V_{\text{sphere}}} = \frac{\frac{\sqrt{3}}{4}\pi R^3 + 2(\frac{16 - 9\sqrt{3}}{24}\pi R^3)}{\frac{4}{3}\pi R^3} = \frac{8 - 3\sqrt{3}}{8} \approx \boxed{35.05 \%}$ .

A Spherical Ring is the solid of revolution of a cyclical segment $AB$ about a diameter $CD$ of its circle, which does not intersect the cyclical segment (figure 1). figure 1 The formula that calculates the volume of a spherical ring is $\boxed{V=\frac{1}{6}\pi A{{B}^{2}}\cdot {A}'{B}'} \ \ \ \ \ (1)$ ,
where ${A}'{B}'$ is the projection of cord $AB$ on the diameter of revolution.

$\ \$

In our case (figure 2),
$A{A}'=\dfrac{R}{2}=\dfrac{OA}{2}$ , thus, $\triangle OA{A}'$ is a $1\text{-}\sqrt{3}\text{-}2$ triangle. Hence, $O{A}'=\frac{\sqrt{3}}{2}R$ , therefore, ${A}'{B}'=2{A}'O=\sqrt{3}R$ .

In addition, $AB$ is parallel to the axis of the hole, i.e. $AB\parallel CD$ , thus, $AB={A}'{B}'=\sqrt{3}R$ . figure 1

Using formula $(1)$ , we have ${{V}_{ring}} =\dfrac{1}{6}\pi {{\left( \sqrt{3}R \right)}^{3}}=\dfrac{\sqrt{3}}{2}\pi {{R}^{3}}$ Finally, the percent of the sphere volume that has been drilled out is $\dfrac{{{V}_{sphere}}-{{V}_{ring}}}{{{V}_{sphere}}}\cdot 100\%=\dfrac{\frac{4}{3}\pi {{R}^{3}}-\frac{\sqrt{3}}{2}\pi {{R}^{3}}}{\frac{4}{3}\pi {{R}^{3}}}\cdot 100\%=\dfrac{8-3\sqrt{3}}{8}\cdot 100\%\approx \boxed{35.05}\%$