The correct answer is $\boxed{\pi}$.
The line integral of a vector field $F$ over a curve $C$ is given by the formula
$$\oint_C F \cdot dr = \int_a^b F(x, y, z) \cdot (dx, dy, dz)$$
where $a$ and $b$ are the endpoints of $C$.
In this case, the vector field $F = yzi$ and the curve $C$ is the circle $x^2 + y^2 = 1$ at $z = 1$. To evaluate the line integral, we can use the parameterization
$$x = \cos t, \quad y = \sin t, \quad z = 1$$
for $0 \leq t \leq 2\pi$. Then,
$$F(x, y, z) = yzi = \sin t \cdot 1 \cdot i = \sin t i$$
and
$$dr = dx \times dy = -\sin t \, dt \times \cos t \, dt = -\sin^2 t \, dt$$
Therefore, the line integral is
$$\oint_C F \cdot dr = \int_0^{2\pi} -\sin^2 t \, dt = \int_0^{2\pi} (1 – \cos 2t) \, dt = \pi$$
The other options are incorrect because they do not take into account the direction of the curve $C$. The curve $C$ is a circle in the counter-clockwise direction, so the line integral must be positive.