= Solution
{c}
First put the nonabelian two-dimensional <Lie algebra> in a convenient form. For any <basis> $p,q$, let $u=[p,q]=ap+bq\ne0$. Then $[p,u]=bu$ and $[q,u]=-au$. Since $(a,b)\ne(0,0)$, some <linear combination> $x$ of $p,q$ has $[x,u]=u$. In particular, $x,u$ are <linearly independent>. Taking $y=u$ gives a <basis> with
$$
[x,y]=y.
$$
This is the <affine Lie algebra of the line>.
Every <alternating bilinear form> on this <vector space> is
$$
\alpha_c(rx+sy,tx+uy)=c(ru-st)\qquad(c\in k).
$$
For any <alternating bilinear form> $\alpha$, the expression
$$
J_\alpha(v,w,t)=\alpha([v,w],t)+\alpha([w,t],v)+\alpha([t,v],w)
$$
is an <alternating trilinear form>: it is trilinear, and setting any two arguments equal makes the terms cancel. Every <alternating trilinear form> on a two-dimensional <vector space> vanishes, since any triple is <linearly dependent>. Hence every $\alpha_c$ satisfies the <Lie algebra two-cocycle> condition, and there are no others:
$$
\boxed{Z^2(\mathfrak s,k)=\{\alpha_c:c\in k\}\cong k,\qquad\alpha_c(x,y)=c.}
$$
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