First, when the two vessels are connected, the water levels equalize at

$$H = \frac{H_1 + H_2}{2} = \frac{10 + 6}{2} = 8\text{ m}.$$

We can get the work done by gravity from the loss of gravitational potential energy:

Initial potential energy (taking the bottom as zero)

For a column of height $H$ and cross-section $A$,

$$U = \rho g A\int_0^H z\,dz = \frac{\rho g A H^2}{2}.$$

So

$$U_i = \frac{\rho g A}{2}\bigl(H_1^2 + H_2^2\bigr) = \frac{10^3\cdot 10\cdot 2}{2}(10^2 + 6^2) = 10^4\,(100 + 36) = 1.36\times10^6\text{ J}.$$

Final potential energy (both at height 8 m)

$$U_f = 2\;\frac{\rho g A H^2}{2}\;\Big|_{H=8} = \rho g A\,(8^2) = 10^3\cdot10\cdot2\cdot64 = 1.28\times10^6\text{ J}.$$

Work done by gravity = drop in potential energy

$$W = U_i - U_f = 1.36\times10^6 - 1.28\times10^6 = 0.08\times10^6 = 8\times10^4\text{ J}.$$

Answer: 8×10^4 J (Option C).