From work energy theorem work done by weight + work done by air $= \Delta KE$

$W_{\large air}=\Delta KE-W_{(\large weight)}$

$ \qquad=\large\frac{1}{2}$$mv^2-mgh$

$\qquad=\large\frac{1}{2} $$ \times 5 \times (10)^2-5 \times 10 \times 20$

$\qquad=-750\;J$

Hence b is the correct answer.