= Solution
The Schwarzschild exterior is vacuum, so
$$
R_{\mu\nu}=0,\qquad R=0,
$$
and consequently $C_{\alpha\beta\gamma\delta}=R_{\alpha\beta\gamma\delta}$. Put
$$
f(r)=1-\frac{2M}{r}.
$$
The relevant <Christoffel symbol>[Christoffel symbols] are
$$
\Gamma^t{}_{tr}=\frac{f'}{2f},
\qquad
\Gamma^r{}_{rr}=-\frac{f'}{2f},
\qquad
\Gamma^r{}_{tt}=\frac12ff'.
$$
With the convention
$$
R^\alpha{}_{\beta\gamma\delta}
=\partial_\gamma\Gamma^\alpha_{\beta\delta}
-\partial_\delta\Gamma^\alpha_{\beta\gamma}+\cdots,
$$
one finds
$$
R_{trtr}=\frac12f''(r).
$$
Since $f''=-4M/r^3$,
$$
\boxed{
C_{trtr}=-\frac{2M}{r^3}}
\qquad(r\gg2M).
$$
This is the leading <Radial Weyl curvature of Schwarzschild spacetime>. The opposite convention for the Riemann tensor reverses the sign but not the physical tidal magnitude.
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