Here is the solution:

Case 1:

Here the weight of wood and lead is balanced by buoyant force on wood and lead.

So wieght of wood and lead = 200 * g + 11.3 * V(Lead) * g        

Bouyant Force ( on Lead and wood as lead is beneath the wood )

= [ V(Wood) + V(Lead) ] * D(Water) * g

As V(wood) = M / D = 200 / 0.8 = 250 cc and D(water) = 1 gm/cc,

Bouyant force = [ 250 + V(Lead) ] * g

As weight = bouyaant force,

that means 200 * g + 11.3 * V(Lead) * g = [ 250 + V(Lead) ] * g

Cancelling 'g', we have 200 + 11.3 * V(lead) = 250 + V(lead)

So V(lead) * (11.3 - 1) = 250 - 200

So V(lead) = 50 / 10.3 = 4.85

So mass of lead = V(lead) * 11.3  = 4.85  * 11.3  = 54.8 gms

Case 2:

Here the weight of wood and lead is balanced by buoyant force on wood only.

So wieght of wood and lead = 200 * g + Mass(Lead) * g        

Bouyant Force ( on wood only as lead is above the wood and lead is not emerged in water )

= V(Wood) * D(Water) * g

As V(wood) = M / D = 200 / 0.8 = 250 cc and D(water) = 1 gm/cc,

Bouyant force = 250 * g

As weight = bouyaant force,

that means 200 * g + Mass(Lead) * g = 250 * g

Cancelling 'g', we have 200 + Mass(Lead) = 250

So Mass(Lead) = 250 - 200 = 50 gm

And so the solution concludes.

Its matches ur answer too.

RATE IF I AM CORRECT