Since $ABCD$ is a square, $\angle BAD=90^\circ$.
Since $ADEFG$ is a regular pentagon, then $\angle DAG$
$\begin{array}{cl}
= & \dfrac{1}{5} \times(5-2)\times 180^\circ \\
= & 108^\circ
\end{array}$
Since $AGHIJK$ is a regular hexagon, then $\angle GAK$
$\begin{array}{cl}
= & \dfrac{1}{6} \times (6-2)\times 180^\circ \\
= & 120^\circ
\end{array}$
In $\Delta ABK$,
$\begin{array}{rcll}
\angle BAK & = & 360^\circ -\angle BAD -\angle DAG -\angle GAK & \text{($\angle$s at a pt)} \\
\angle BAK & = & 360^\circ-90^\circ-108^\circ-120^\circ \\
\angle BAK & = & 42^\circ
\end{array}$
Since $AB=AK$, then we have
$\begin{array}{rcll}
\angle ABK & = & \angle AKB & \text{(base $\angle$s, isos. $\Delta$)}
\end{array}$
Hence, we have
$\begin{array}{rcll}
\angle ABK & = & \dfrac{1}{2}(180^\circ -\angle BAK) & \text{($\angle$ sum of $\Delta$)} \\
\angle ABK & = & \dfrac{1}{2}(180^\circ-42^\circ) \\
\angle ABK & = & 69^\circ
\end{array}$
