The statement in question refers to the Kudo-Serre transgression theorem: If $E_r$ is, for example, the LHS spectral sequence of a group extension and $x \in E_{2k+1}^{0,2k}$ is transgressive with $d_{2k+1}(x)=y$, then $$d_i(x^p)=0\;\;(i=2k+1,...,2kp)\quad \text{and }\;\; d_{2kp+1}(x^p)=P^ky$$ where $P^k$ denotes the Steenrod power operation (a textbook reference is McCleary: A User's guide to spectral sequences, Theorem 6.14).

I'll give an example that is related to the Benson-Feshbach paper: Let $p$ be a prime and let $$1 \to C \to G \to Q \to 1$$ be a central extension with $C \cong (\mathbb{Z}/p)^n$ elementray abelian. If $k$ is a field of char. $p$ (with trivial $G$-action), than $Q$ acts trivially on $H^\ast(C;k)$ and by the universal coefficient theorem, the LHS spectral sequence satisfies
$$E_2^{ij} \cong H^i(Q;k) \otimes_k H^j(C;k).$$
Moreover, $E_2 \cong H^\ast(Q;k) \otimes_k H^\ast(C;k)$ is an isomorphism of $k$-algebras (up to a sign). This is what is ment by B-F when they say the $E_2$-page is easy to compute.

The cohomology of $C$ is given by $H^\ast(C;k) = k[x_1,...,x_n,y_1,...,y_n]$ with $\deg(x_i)=1, \deg(y_j)=2$ and the relations $x_i^2=y_i$ if $p=2$ and $x_i^2=0$ for $p$ odd. Note that $y_i = \beta(x_i)$ where $\beta$ is the Bockstein homomorphism.

In terms of the spectral sequence, we have $x_i \in E_2^{0,1}$. Hence $d_2(x_i) =: z_i \in E_2^{2,0}=H^2(Q;k)$, i.e. $x_i$ is transgressive. Now Kudo's theorem says that $d_2, \beta$ commute, i.e.

$$d_2(y_i) = d_2(\beta(x_i))=\beta(d_2(x_i))=\beta(z_i) \in H^3(Q;k).$$
Hence $y_i$ is again transgressive and by Kudo's theorem $d_5(y_i^2)=P^1\beta(z_i).$ Continuing this way, one obtains
$$d_{2p^k+1}(y_i^{p^k})=P^{p^{k-1}}P^{p^{k-2}}\cdots P^1\beta(z_i).$$
As a result, the differentials on the fiber $E_\ast^{0,\ast}$ are completely determined by the values of $z_i \in H^2(Q;k)$ (the base) under the power operations.