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$W_1$= Work done against gravity $=mgh$

$ \qquad=V \rho gh$

Work done against pressure difference

$W_2=\Delta PV= h \rho gV$

Total work done $=W_1 +W_2$

$\qquad= 2h \rho gV$

$P= \large\frac{W}{t}$

$\quad= \large\frac{2 \times 6 \times 10^3 \times 10 \times 300 \times 10^{-3}}{60}$

$\quad= 600\; walt$

Hence a is the correct answer.

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