A cubical block of wood weighing 200 g has a lead piece fastened...

Let the mass of the lead piece be $m$.

Volume of the wooden block, $V_w = \dfrac{200}{0.8} = 250 \text{ cm}^3$

Volume of the lead piece, $V_l = \dfrac{m}{11.3} \text{ cm}^3$

For the block to just float in water, the entire system (wood and lead) must be completely submerged. By the principle of flotation, the total mass of the system equals the mass of the displaced water.

$200 + m = \left(V_w + V_l\right) \times \rho_{\text{water}}$

Substituting the values, with $\rho_{\text{water}} = 1 \text{ g/cm}^3$:

$200 + m = \left(250 + \dfrac{m}{11.3}\right) \times 1$

$m - \dfrac{m}{11.3} = 250 - 200$

$m \left(\dfrac{11.3 - 1}{11.3}\right) = 50$

$m \left(\dfrac{10.3}{11.3}\right) = 50$

$m = \dfrac{50 \times 11.3}{10.3} = \dfrac{565}{10.3} \approx 54.8 \text{ g}$