1. Interpret the setup

A uniform rigid rod of length $27\,\text{cm}$ has two masses hung at its ends:

• one end: unknown cube of side $10\,\text{cm}$
• other end: $200\,\text{g}$ mass

The rod is supported on a wedge (pivot point).

Given: distance between the wedge point and the $200\,\text{g}$ mass is $25\,\text{cm}$. Since the rod length is $27\,\text{cm}$, the distance between wedge and the unknown mass end is $27-25=2\,\text{cm}.$

So the lever arms are:

• unknown mass side: $2\,\text{cm}$
• $200\,\text{g}$ side: $25\,\text{cm}$

2. Condition at balance

Initially it was not balanced. Water is added below the unknown cube, and at balance, half of the cube is submerged.

Since the rod is rigid and uniform, its own weight acts at its center. But because the rod is uniform, its center is at $13.5\,\text{cm}$ from either end.

Let us check pivot position:

• pivot is $25\,\text{cm}$ from the $200\,\text{g}$ end
• hence pivot is $2\,\text{cm}$ from the unknown-mass end

So the rod's center is $13.5\,\text{cm}$ from the unknown end, i.e. at a distance $13.5-2=11.5\,\text{cm}$ from the pivot toward the $200\,\text{g}$ side.

Thus rod's weight would also contribute torque. However, the problem does not provide rod mass, so the only consistent interpretation is that the support point is chosen such that rod's own effect is irrelevant/neglected compared to the attached masses, i.e. balance is to be written using the end loads only.

Hence at equilibrium: $\text{effective weight of cube}\times 2 = 200\times 25$ where masses are in grams-force units.

So, $W_{\text{eff}}\cdot 2 = 5000$ $W_{\text{eff}} = 2500\,\text{g}$

Thus the cube behaves as if its downward force is equivalent to $2500\,\text{g}$.

3. Find buoyant force on the cube

Cube side $=10\,\text{cm}$, so volume is $V=(10)^3=1000\,\text{cm}^3.$

Half volume submerged means displaced water volume is $V_{\text{sub}}=\frac{1000}{2}=500\,\text{cm}^3.$

Since density of water is $1\,\text{g/cm}^3$, buoyant force equals weight of displaced water: $B = 500\,\text{g-force}.$

So if actual mass of cube is $M$ grams, its effective weight in water is $M-500.$

At balance, this equals $2500\,\text{g}$: $M-500=2500$ $M=3000\,\text{g}=3\,\text{kg}.$

4. Final answer

The unknown mass is $\boxed{3\,\text{kg}}.$