\[
\begin{array}{|l|l|l|l|l|}
\hline
g(x) & y & \operatorname{sign}(g(x)) & \text{0-1 loss} & \text{Squared loss } [g(x)-y]^2 \\
\hline
+30 & +1 & +1 & 0 & 29^2 \\
\hline
-20 & -1 & -1 & 0 & 19^2 \\
\hline
-1 & -1 & -1 & 0 & 0 \\
\hline
+1 & +1 & +1 & 0 & 0 \\
\hline
\end{array}
\]
The value of $0-1$ loss is zero, while the value of squared loss is a large positive quantity.
Hence, option (C) is correct.