Answer
$218\;\rm cycle$
Work Step by Step
Let's assume that the soda is water which means that its density is 1000 kg/m$^3$.
Now we need to find the total heat exhausted from this soda,
$$Q=mc\Delta T$$
where $m=\rho V$,
$$Q= \rho Vc_{water}\Delta T\tag 1$$
Now we need to find the amount of heat exhausted in one cycle in a Carnot cycle,
$$\eta=1-\dfrac{T_C}{T_H}=1-\dfrac{Q_C}{Q_H}$$
Hence,
$$\dfrac{Q_C}{Q_H}=\dfrac{T_C}{T_H}$$
$$Q_C=Q_H\left[\dfrac{T_C}{T_H}\right]\tag 2$$
The number of cycles needed to cool the soda down from 25$^\circ$C to 5$^\circ$C is given by
$$N=\dfrac{Q}{Q_C}$$
Plugging from (1) and (2);
$$N=\dfrac{ \rho Vc_{water}\Delta T}{Q_H\left[\dfrac{T_C}{T_H}\right]}$$
Plugging the known;
$$N=\dfrac{(1000) (500\times 10^{-6})(4190)(25-5)}{250\left[\dfrac{-20+273}{55+273}\right]}=217.3$$
$$N=\color{red}{\bf 218}\;\rm cycle$$
