The relationship between the differential scattering cross section $d \sigma/ d \Omega$ , the solid angle $d \Omega$ , the number of incident particles per unit time $L$ , the number of incident particles scattered into the solid angle $d \Omega$ per unit time $d N$ , and number of nuclei per unit area $n$ is %(see Griffiths QM page 399 but note that the notation there is slightly different)
$\begin{align*}\frac{d \sigma}{d \Omega} L n d \Omega = d N\end{align*}$
We are given that
$\begin{align*}L &= 10^{12}\mbox{ 1/s} \\n &= 10^{20} \;\mathrm{cm}^{-2} \\dN &= 10^2 \mbox{ 1/s} \\d \Omega &...$
Plugging these values in, we find that
$\begin{align*}\frac{d \sigma}{d \Omega} = \frac{d N}{L n d \Omega} = \boxed{10^{-26} \;\mathrm{cm}^2\mathrm{/}\mathrm{steradi...$
Therefore, answer (C) is correct.

Solution to 1992 Problem 95