1. Identify the pivot point

The rod is balanced on a sharp edge at the 40 cm mark. So, take moments about the 40 cm mark.

2. Locate all forces and their distances from the pivot

• Mass of rod = $250\,\text{g}$, uniform, so its weight acts at the center i.e. at $50\,\text{cm}$ mark.
• A mass of $400\,\text{g}$ is suspended at $10\,\text{cm}$ mark.
• Let the required mass at $90\,\text{cm}$ mark be $m$ grams.

Distances from pivot ($40\,\text{cm}$ mark):

• Rod’s center: $50-40=10\,\text{cm}$ to the right
• $400\,\text{g}$ mass: $40-10=30\,\text{cm}$ to the left
• Unknown mass at $90\,\text{cm}$: $90-40=50\,\text{cm}$ to the right

3. Apply condition of rotational equilibrium

For balance, clockwise moment = anticlockwise moment.

• Left side moment due to $400\,\text{g}$: $400 \times 30$
• Right side moments:
  • rod: $250 \times 10$
  • unknown mass: $m \times 50$

So, $400 \times 30 = 250 \times 10 + m \times 50$

4. Solve for $m$

$12000 = 2500 + 50m$

$50m = 12000 - 2500 = 9500$

$m = \frac{9500}{50} = 190\,\text{g}$

5. Check options

The required mass is: $\boxed{190\,\text{g}}$

So the correct option is C.