A lead piece is fastened to the top surface of a cubical block of...

Let the mass of the wood block be $m_w = 200 \text{ g}$.

The specific gravity of wood is $0.8$, so its density is $\rho_w = 0.8 \text{ g/cm}^3$.

The volume of the wood block is:
$V_w = \dfrac{m_w}{\rho_w} = \dfrac{200}{0.8} = 250 \text{ cm}^3$

The block floats with its upper surface just dipping into the water. This means the entire volume of the wood block is submerged in the water. Since the lead piece is fastened to the top surface, it remains above the water level and does not displace any water.

The buoyant force is provided entirely by the submerged wood block. According to Archimedes' principle, the buoyant force equals the weight of the displaced water:
$F_B = V_w \times \rho_{\text{water}} \times g = 250 \times 1 \times g = 250g \text{ dynes}$

The total downward force is the combined weight of the wood block and the lead piece:
$W_{\text{total}} = (m_w + m_{\text{lead}})g = (200 + m_{\text{lead}})g$

For the block to float, the total weight must equal the buoyant force:
$(200 + m_{\text{lead}})g = 250g$

$200 + m_{\text{lead}} = 250$

$m_{\text{lead}} = 50 \text{ g}$

The specific gravity of lead ($11.3$) is not required in this case since the lead piece is not submerged.

Answer: $50 \text{ g}$