Realizing $\mathcal{A}(2)//\mathcal{A}(1)$ by a finite spectrum Let $\cal A$ denote the mod 2 Steenrod algebra. Can the $\mathcal{A}(2)$-module structure on $\mathcal{A}(2)//\mathcal{A}(1)$ be enriched to an $\cal A$-module structure? If so, is there a finite spectrum $X$ such that $H^*(X) = \mathcal{A}(2)//\mathcal{A}(1)$?
 A: As pointed out by John Rognes, this answer is not correct (it misses the $Sq^4$ from the bottom class to the class in dimension $4$). Sorry for the confusion.
=========== previous answer ================================
The compact real Lie group $G_2$ realizes the desuspension of $A(2)/\!/A(1)$. Recall that
$$H^\ast(G_2;{\mathbb F}_2) = {\mathbb F}_2[x_3,x_5]/(x_3^4,x_5^2)$$
with $Sq^2x_3=x_5$, $Sq^1x_5 = x_3^2$. If I'm not mistaken you get an isomorphism to $A(2)/\!/A(1)$ via 
$$x_3\leftrightarrow Sq(4),\quad x_5\leftrightarrow Sq(0,2),\quad x_3^2\leftrightarrow Sq(0,0,1).$$
A: The $A(2)$-module structure on $A(2)//A(1)$ does not extend to an $A$-module structure.  In particular, there is no spectrum $X$ with $H^*(X; F_2) = A(2)//A(1)$ as an $A(2)$-module.
Additively, $A(2)//A(1)$ is generated by classes $g_i$ in degree $i$
for $i = 0, 4, 6, 7, 10, 11, 13$ and $17$.  The Adem relation $Sq^4 Sq^6 = Sq^{10} + Sq^8 Sq^2$ implies $Sq^{10}(g_0) = g_{10}$.  The Adem relation $Sq^2 Sq^8 = Sq^{10} + Sq^9 Sq^1$ implies $Sq^{10}(g_0) = 0$.  This contradicts the existence of any $A$-module structure extending the given $A(2)$-module structure.
PS: Bruner's ext code (http://www.math.wayne.edu/~rrb/papers/) contains a script (newconsistency) that lets you verify that a purported presentation really defines an $A(2)$-module, and tells you what is needed to extend the presentation to an $A$-module structure. It could be handy if you want to realize other $A(2)$-modules by spectra.
