Answer
The missing values are $B=9{}^\circ,C=148{}^\circ \text{ and }c\approx 73.6$.
Work Step by Step
For any triangle,
The law of sines states that:
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\operatorname{sinC}}$
The law of cosines states that:
$\begin{align}
& {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2\cdot b\cdot c\cdot \cos A \\
& {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2\cdot a\cdot c\cdot \cos B \\
& {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2\cdot a\cdot b\cdot \cos C \\
\end{align}$
Now from the law of sines we get:
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{54.3}{\sin 23{}^\circ }=\frac{22.1}{\sin B} \\
& 54.3\left( \sin B \right)=22.1\left( \sin 23{}^\circ \right) \\
& \sin B=\frac{22.1\left( \sin 23{}^\circ \right)}{54.3}
\end{align}$
We will simplify it further to obtain the measure of angle B.
$\begin{align}
& \sin B=0.15 \\
& B={{\sin }^{-1}}\left( 0.15 \right) \\
& B\approx 9{}^\circ
\end{align}$
Now using the angle sum property:
$\begin{align}
& A+B+C=180{}^\circ \\
& 23{}^\circ +9{}^\circ +C=180{}^\circ \\
& C=180{}^\circ -32{}^\circ \\
& C=148{}^\circ
\end{align}$
Finally, from the law of sines:
$\begin{align}
& \frac{a}{\sin A}=\frac{c}{\sin C} \\
& \frac{54.3}{\sin 23{}^\circ }=\frac{c}{\sin 148{}^\circ } \\
& c\cdot \sin 23{}^\circ =54.3\left( \sin 148{}^\circ \right) \\
& c=\frac{54.3\left( \sin 148{}^\circ \right)}{\sin 23{}^\circ }
\end{align}$
This gives $c\approx 73.6$.
So, $B=9{}^\circ,C=148{}^\circ \text{ and }c\approx 73.6\text{ }$.
