This is a surprisingly simple thing to calculate.
It is a well known result that a consequence of the inverse square law is that there is no force inside a symmetrical hollow shell. This means that as the object falls into the hole, it will appear to be attracted by a sphere of decreasing radius - the mass outside "doesn't count."
The acceleration of gravity at the surface of a sphere of radius R (assuming uniform density $\rho$) is given by
$$\begin{align}
a &= \frac{GM}{R^2} \\
&= \frac{4G\pi R^3\rho}{3R^2} \\
&= \frac43\pi \rho G R\\
\end{align}$$
Where $G$ is the gravitational constant, and $R$ is the distance to the center of the earth. In other words - the acceleration is proportional to the distance to the center. The corollary is that an object dropped into a hole through the center of the earth will exhibit simple harmonic motion.
Let's do the math in more detail. Put the distance from the center as $r$, then the acceleration $\frac{d^2r}{dt^2}$ is given by
$$\frac{d^2r}{dt^2}=-\frac43 \pi \rho G r$$
(since the acceleration is pointing towards the center). This looks like the differential equation for simple harmonic motion:
$$\frac{d^2x}{dt^2} = -\omega^2 x$$
for which the solution (if velocity is zero at t=0, and amplitude is $x_0$) is
$$x(t) = x_0 \cos\omega t$$
and the velocity is
$$v(t) = -\omega x_0 \sin\omega t$$
It follows that we can write the expression for the velocity as a function of position by eliminating time:
$$\begin{align}\\
v(x) &= -\omega x_0 \sin \cos^{-1}\left(\frac{x}{x_0}\right)\\
&= -\omega x_0\sqrt{1-\left(\frac{x}{x_0}\right)^2}\end{align}$$
If we now substitute $\omega^2 = \frac43 \pi \rho G = \frac{g}{R}$ where $g$ is the gravitational acceleration at the surface of the earth, we get
$$\begin{align}\\
v(r) &= -\sqrt{\frac{g}{R}} R \sqrt{1-\left(\frac{r}{R}\right)^2}\\
&=\sqrt{\frac{g}{R}\left(R^2-r^2\right)}
\end{align}$$
If we substitute $r=R$, we get $v=0$ as expected; and when we put in $r=0$, we get $v = \sqrt{gR}$ = 7904 m/s using your values for $R$ and $g$. I think that's pretty close...
Best Answer
The geometry of spacetime is described by a function called the metric tensor. If you're starting to learn GR then any moment you'll encounter the Schwarzschild metric that describes the geometry outside a sphrically symmetric body. When you go inside the body the geometry is described by the (less well known) Schwarzschild interior metric.
The exact form of the interior metric depends on how the density of the spherical body changes with depth, so for real objects is has a rather complicated form. However for a body with constant density it simplifies a bit and looks like:
$$ ds^2 = -\left[\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\sqrt{1-\frac{2Mr^2}{R^3}}\right]^2dt^2 + \frac{dr^2}{\left(1-\frac{2Mr^2}{R^3}\right)} + r^2 d\Omega^2 \tag{1} $$
In this equation $r$ is the distance from the centre of the object with mass $M$, and $R$ is the radius of the object. So this metric applies for $r \le R$.
If you ignore relativity for a moment and consider just Newtonian gravity, then the radial acceleration of an object near a spherical body is:
$$ a = \frac{d^2r}{dt^2} = -\frac{GM}{r^2} $$
In GR the situation is rather more complicated (which I'm sure comes as no surprise) but we can define an analogous quantity called the four-acceleration. In particular we want the radial component $d^2r/d\tau^2$. Although this looks like the Newtonian acceleration above, it's the second derivative with respect to proper time $\tau$ not the coordinate time $t$.
We get the four acceleration using the geodesic equation:
$$ {d^2 x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} $$
where the $\Gamma^\mu_{\alpha\beta}$ are called Christoffel symbols (of the second kind) and they depend on the spacetime curvature. This all looks rather complicated, but for spherically symmetric objects all but one of the sixteen terms on the right hand side are zero and the radial four acceleration of a stationary object is simply:
$$ \frac{d^2r}{d\tau^2} = - \Gamma^r_{tt} u^t u^t \tag{2} $$
This is the equation that tells us how the (four) acceleration varies with depth, and we can use it to show that you are weightless at the centre of the object. Equation (2) is going to go to zero if either $u^t$ becomes zero or $\Gamma^r_{tt}$ becomes zero. The quantity $u^t$ is the time component of the four-velocity, which you can think of as the time dilation factor. This is simply $dt/d\tau$ for constant $r$, $\theta$ and $\phi$, and we get it from the metric (1) simply by setting $dr = d\theta = d\phi = 0$ to give:
$$ ds^2 = -\left[\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\sqrt{1-\frac{2Mr^2}{R^3}}\right]^2dt^2 $$
Since $ds^2 = -d\tau^2$ a simple rearrangement gives:
$$ u^t = \frac{dt}{d\tau} = \frac{1}{\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\sqrt{1-\frac{2Mr^2}{R^3}}} $$
Let's just note that this does not go to zero as $r \rightarrow 0$, and move swiftly on to look at the other term $\Gamma^r_{tt}$. Calculating this involves some painful algebra, but Mathematica is good at this sort of thing and Danu helpfully used Mathematica to do the calculation for me. The result is:
$$\Gamma_{tt}^r= \frac{r}{2R^6}\left[2M^2r^2+MR^3\left(3\sqrt{1-\frac{2Mr^2}{R^3}}\sqrt{1-\frac{2M}{R}}-1\right)\right] $$
Yet another fiendishly complicated expression, but note that the whole thing is multiplied by $r/2R^6$ and that means if $r = 0$ the whole complicated expression is just zero.
And there's our result!
When $r = 0$ the Christoffel symbol $\Gamma_{tt}^r$ is zero and that means the radial four-acceleration is zero and that means you're weightless.