Activity of Polonium in half-time in Becquerels
Well, first of all, I should point out that your answer and the would be the answer ($1.7$ times your answer) are both wrong by the fact that after 552 days (4 half-lives), the activity should be $\displaystyle\frac{\pu{2E6 MBq}}{16} = \pu{1.25E5 MBq}.$ Since this is smaller than both your answer and the would be answer, they both would be incorrect. Accordingly, correct answer would be smaller than $\pu{1.25E5 MBq},$ because the sample would be further decays for another six days. Let’s do the proper calculations:
The half-life $\left(t_\frac{1}{2}\right)$ of a radioactive element is defined as
$$t_\frac{1}{2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}.\label{eqn:1}\tag{1}$$
The $t_\frac{1}{2} = \pu{138 d}$ in the equation \eqref{eqn:1} for your case. Thus from the equation \eqref{eqn:1}, you can calculate for $\lambda:$
$$\lambda = \frac{\ln 2}{t_\frac{1}{2}} = \frac{0.693}{\pu{138 d}} = \pu{5.02E-3 d-1}. \tag{2}$$
Also, the radioactive decay is according to the equation:
$$A_t = A_0\mathrm e^{-\lambda t}.\label{eqn:3}\tag{3}$$
where $A_0$ is initial activity and $A_t$ is the activity of the same same sample after time $t$ (Note: $t_\frac{1}{2}$ and $t$ should have same units). Thus, applying the $\lambda$ value on the equation \eqref{eqn:3}:
$$A_t = \pu{2E6 MBq}\times\exp\left(-\pu{5.02E-3 d-1}\times\pu{558 d}\right) = \pu{1.21E5 MBq}.$$
Note: I disregard the correct significant figures because I believe the initial activity should be given in three significant figures.
For fun, you can also checked the validity of the answer using my initial remark. You can use the activity of the sample after four consecutive $t_\frac{1}{2}$s (after $138 \times 4 = \pu{552 d})$ as its initial activity and find the activity after another six days using the equation \eqref{eqn:3}:
$$A_t = \pu{1.25E5 MBq}\times\exp\left(-\pu{5.02E-3 d-1}\times\pu{6 d}\right) = \pu{1.21E5 MBq}.$$