One direction: 6(a). The other direction: if a normal space is an absolute retract then it has the universal extension property. The idea, as I understand, is as follows: we consider a continuous function $f$ from a closed subspace $A$ of $X$ to $Y$ ; we know that whenever $Y$ is homeomorphic to a closed subspace of a normal space, the subspace is a retract, i.e. there exists a continuous map from the space onto the subspace that preserves all points of the subspace; but we need a continuous function from $X$ to $Y$ that extends $f$ ; we consider the space which is the union of the spaces $X$ and $Y$ and group some points into equivalence classes such that every point in $Y$ belongs to a separate class that also contains the preimage of the point; the union of all classes for all points in $Y$ is a closed subspace of the quotient space being homeomorphic to $Y$ ; we show the quotient space is normal and consider a retraction from $X$ onto the set of classes of points in $Y$ ; now we can construct a continuous extension of $f$ , as we shall see.Construct the adjunction space $Z_f$ as hinted: the quotient space obtained from $Z=X\cup Y$ by identifying each point $y\in Y$ with points in $f^{-1}(y)$ (for this we, first, need to specify the topology on $Z$ : a set $U$ is open in $Z$ iff $U\cap X$ is open in $X$ and $U\cap Y$ is open in $Y$ ). Let $p$ be the quotient map. Let $B=X-A$ and denote elements of $Z_f=[B]\cup[Y]$ as follows: $[x]\in[B]$ where $x\in B$ and $[y]\in[Y]$ where $y\in Y$ (if $y\notin f(A)$ then $[y]$ contains only one element).First, we show that $Z_f$ is normal. Let $[S_0]$ and $[S_1]$ be closed in $Z_f$ . Let $[T_i]=[S_i]\cap[Y]$ and $[R_i]=[S_i]\cap[B]$ . $[T_i]$ are closed in $[Y]$ ($[Y]$ itself is closed as it contains all elements from $Y$ and $A$ , both closed in $Z$ ), therefore, $T_i$ are closed in $Y$ and we can separate them by a continuous function $g(T_i)=i$ (Urysohn lemma). Now, $g'=g\circ f$ is continuous on $A$ and $g'(f^{-1}(T_i))=i$ . We can extend it onto $R_i$ : $g'(R_i)=i$ . It is still continuous: it is continuous on $R_i\cup f^{-1}(T_i)$ and $A$ (the pasting lemma?). Now we can extend it onto $X$ (Tietze theorem). Note that the extended $g'$ is constant on $p^{-1}([z])$ for all $z\in Z$ . Therefore, it induces a continuous function $g"$ on $Z_f$ (Section 22) such that it separates $[S_0]$ and $[S_1]$ . Hence, $Z_f$ is normal.Now, we show that $Y$ is homeomorphic to a closed subset of $Z_f$ , namely, to the set $[Y]$ . Indeed, $h=p|Y:Y\to[Y]$ is continuous and bijective. It maps closed $T\subseteq Y$ to $[T]=p(T\cup f^{-1}(T))$ which is the image of a closed saturated subset of $Z$ under the quotient map ($f^{-1}(T)$ is closed in $A$ which is closed in $X$ ).Finally, since $[Y]$ is a closed subset of the normal space $Z_f$ homeomorphic to $Y$ , and $Y$ is an absolute retract, there is a retraction $r:Z_f\to[Y]$ such that for every $y\in Y$ : $r([y])=[y]$ . Let $f'=h^{-1}\circ r\circ p$ . $f'$ is continuous, and maps $X\cup Y$ to $Y$ . For every $a\in A$ : $f'(a)=h^{-1}(r([f(a)]))=h^{-1}([f(a)])=f(a)$ . Therefore, $f'|X$ extends $f$ .