Outer radius $R = 6$ cm, inner radius $r = 6 - 1 = 5$ cm.
Volume of steel = Volume of outer hemisphere − Volume of inner hemisphere
$$= \frac{2}{3}\pi R^3 - \frac{2}{3}\pi r^3 = \frac{2}{3}\pi(6^3 - 5^3) = \frac{2}{3}\pi(216 - 125) = \frac{2}{3}\pi \times 91 = \frac{182}{3}\pi \text{ cm}^3$$
Answer: (B) $\dfrac{182}{3}\pi$
The key idea: the bowl is hollow, so subtract the volume of the inner hemisphere from the outer one. Thickness = 1 cm gives inner radius = outer radius − 1 = 5 cm. Use $V = \frac{2}{3}\pi r^3$ for a hemisphere. Many students mistakenly use full sphere formula — remember it's a hemispherical bowl, so use $\frac{2}{3}\pi r^3$.