A cylindrical rod with one end in a steam chamber and the other end...

The rate of heat flow through a rod is given by $H = \dfrac{KA\Delta T}{L}$.

The rate of melting of ice is $m = \dfrac{H}{L_f} = \dfrac{KA\Delta T}{L L_f}$, where $L_f$ is the latent heat of fusion.

For the first rod, the rate of melting is:

$m_1 = \dfrac{K_1 \pi r_1^2 \Delta T}{L_1 L_f} = 0.1 \text{ g/s}$

For the second rod, the given parameters are $K_2 = \dfrac{K_1}{4}$, $r_2 = 2r_1$, and $L_2 = \dfrac{L_1}{2}$.

The rate of melting for the second rod is:

$m_2 = \dfrac{K_2 \pi r_2^2 \Delta T}{L_2 L_f}$

Substituting the values of $K_2$, $r_2$, and $L_2$:

$m_2 = \dfrac{\left(\dfrac{K_1}{4}\right) \pi (2r_1)^2 \Delta T}{\left(\dfrac{L_1}{2}\right) L_f}$

m_2 = \dfrac{\left(\dfrac{K_1}{4}\right)^{4} \pi r_1^2 \Delta T}{\dfrac{L_1}{2} L_f}$<br><br>$m_2 = \dfrac{K_1 \pi r_1^2 \Delta T}{\dfrac{L_1}{2} L_f} = 2 \left( \dfrac{K_1 \pi r_1^2 \Delta T}{L_1 L_f} \right)

$m_2 = 2 m_1$

$m_2 = 2$\times0.1 = 0.2 \text{ g/s}

Answer: 0.2