Answer
$98\;\rm cycle$
Work Step by Step
We know that the thermal efficiency of the Carnot engine is given by
$$\eta_{\rm Carnot}=1-\dfrac{T_C}{T_H}$$
which is also given by
$$\eta_{\rm Carnot}=\dfrac{W_{\rm out}}{Q_H}$$
Thus,
$$1-\dfrac{T_C}{T_H}=\dfrac{W_{\rm out}}{Q_H}$$
Thus the work done in one cycle is given by
$${W_{\rm out}}=Q_H\left[1-\dfrac{T_C}{T_H}\right]\tag 1$$
And the total work needed to raise the 10-kg mass a 10 m up is given by
$$W=F_g\Delta y=mgh\tag 2$$
Hence, the number of cycles needed to raise the mass is given by
$$N=\dfrac{W}{{W_{\rm out}}}$$
Plugging from (1) and (2);
$$N=\dfrac{mgh}{Q_H\left[1-\dfrac{T_C}{T_H}\right]}$$
Plugging the known;
$$N=\dfrac{(10)(9.8)(10)}{(25)\left[1-\dfrac{0+273}{182+273}\right]}$$
$$N=\color{red}{\bf 98}\;\rm cycle$$
