Answer
No such triangle.
Work Step by Step
Using the Sine Law:
$\dfrac{\sin A}{a} = \dfrac{\sin B}{b} \\
\dfrac{\sin 45.6^\circ}{234} = \dfrac{\sin B}{567}\\
\sin B = \dfrac{567 \sin 45.6^\circ}{234} \\
\sin B \approx 1.7312
$
Thus, for such values of $a,b$, and $A$, the Sine Law gives $\sin B \approx 1.7312 > 1$.
Since the sine of any angle cannot exceed $1$, no such triangle exists.
