The locus of point of intersection of the tangents to the circle $x^2+y^2=16$, such that the angle between them is $60^{\circ}$, is

The circle $x^2 + y^2 = 16$ has center at the origin and radius $4$. For two tangents from an external point $P(h, k)$ intersecting at $60^\circ$, the line from the center to $P$ bisects the angle between the tangents, making the angle between $OP$ and either tangent $30^\circ$.

In right triangle $\triangle OTP$, where $T$ is the point of tangency, $\sin 30^\circ = \frac{r}{OP} = \frac{4}{OP}$. Since $\sin 30^\circ = \frac{1}{2}$, we have $OP = 8$.

The distance $OP = \sqrt{h^2 + k^2} = 8$, so squaring both sides yields $h^2 + k^2 = 64$. Replacing $(h, k)$ with $(x, y)$ gives the locus $x^2 + y^2 = 64$.

Final answer: $\boxed{\text{B}}$