1. Understand the geometry

The two conducting walls are parallel plates of dimensions $50\text{ cm} \times 50\text{ cm}$ so the effective plate area is $A = 50 \times 50 = 2500\text{ cm}^2 = 0.25\text{ m}^2.$

The separation between the plates is the width of the container: $d = 5\text{ cm} = 0.05\text{ m}.$

The total height of the container is $50\text{ cm}.$

2. Find the volume of liquid filled in 10 s

Liquid is poured at the rate $250\text{ cm}^3\text{/s}.$

So in $10$ s, volume filled is $V = 250 \times 10 = 2500\text{ cm}^3.$

3. Find the height of liquid

Base area of container: $A_{\text{base}} = 50\text{ cm} \times 5\text{ cm} = 250\text{ cm}^2.$

Hence liquid height after 10 s is $h = \frac{V}{A_{\text{base}}} = \frac{2500}{250} = 10\text{ cm}.$

So:

• lower $10$ cm region between plates contains dielectric of constant $K=3$,
• upper $40$ cm region contains air $(K=1)$.

4. Model as two capacitors in parallel

Since the dielectric fills the lower part between the same pair of plates, the arrangement is equivalent to two parallel capacitors:

• one with dielectric, area $A_1$
• one with air, area $A_2$

Areas: $A_1 = 50\text{ cm} \times 10\text{ cm} = 500\text{ cm}^2 = 0.05\text{ m}^2,$ $A_2 = 50\text{ cm} \times 40\text{ cm} = 2000\text{ cm}^2 = 0.20\text{ m}^2.$

Total capacitance: $C = \frac{K\epsilon_0 A_1}{d} + \frac{\epsilon_0 A_2}{d}.$

Substitute values: $C = \frac{\epsilon_0}{d}(3A_1 + A_2).$

Now, $3A_1 + A_2 = 3(0.05) + 0.20 = 0.15 + 0.20 = 0.35\text{ m}^2.$

Therefore, $C = \frac{9\times 10^{-12}}{0.05}\times 0.35.$

$C = 9\times 10^{-12} \times 7.$

$C = 63\times 10^{-12}\text{ F} = 63\text{ pF}.$

5. Check options

The correct option is: $\boxed{\text{B: }63\text{ pF}}$